Solutions to Exercises from John S. Rose's _A Course on Group Theory_

--- google-site-verification: 78Z5MJBzbSoZRKem4sXkuUCuaAXo3OXaq2AKjX8z3h0 --- # Solutions to Exercises from John S. Rose's _A Course on Group Theory_ ###### tags: `Group Theory`, `Groups`, `Group Actions`, `Solutions` <style> .markdown-body { background-color: #FDFDFA; color: #102020; box-shadow: 0 4px 6px 0 rgba(0, 0, 0, 0.2), 0 4px 6px 0 rgba(0, 0, 0, 0.19); } p { font-family: 'Tahoma'; text-align: justify; hyphens: auto; margin: 0px 18px 5px 2px; } .q { font-family: 'Verdana', sans-serif; } ol { margin-left: 10px; } h1, h2 { font-variant: small-caps; text-shadow: 1px 1px #F0F0F0; margin: 0px 50px 5px 15px; } h6 { margin: 0px 20px 5px 15px; } </style> $\DeclareMathOperator{\lcm}{lcm}$ $\DeclareMathOperator{\Im}{Im}$ $\DeclareMathOperator{\Hom}{Hom}$ $\DeclareMathOperator{\GL}{GL}$ $\DeclareMathOperator{\Aut}{Aut}$ $\DeclareMathOperator{\Inn}{Inn}$ $\DeclareMathOperator{\Isom}{Isom}$ $\DeclareMathOperator{\Rot}{Rot}$ ## Chapter 0 1. Any group of prime order is cyclic. ***Sol.*** The only divisors of a prime number are $1$ and the number itself, and therefore the order of any subgroup must be equal to $1$ or the order of the whole group. Thus, the only non-trvial subgroup is the whole group itself. Now, any non-identity element of the group (which must exist, since $2$ is the least prime number) generates a non-trivial cyclic subgroup, which is necessarily equal to the whole group. 2. Any two cyclic groups of the same finite order are isomorphic. ***Sol.*** Let $G = \langle g \rangle$ and $H = \langle h \rangle$ be two cyclic groups of finite order $n.$ Define $f \colon G \to H,$ by $f(g^i) = h^i,$ for all $g \in G.$ Then $f$ is well-defined and injective, since $g^i = g^j$ if and only if $i \equiv j \mod n,$ which is equivalent to $h^i = h^j.$ For any two elements $g^i, g^j \in G,$ $f(g^i g^j) = f(g^{i + j}) = h^{i + j} = h^i h^j = f(h^i) f(h^j).$ Hence, $f$ is an injective homomorphism from $G$ to $H,$ and therefore is an isomorphism, since $|G| = |H|.$ 3. If $g^2 = 1$ for every $g \in G,$ then $G$ is an Abelian group. ***Sol.*** For all $g \in G,$ $g^2 = 1 \implies g = g^{-1}.$ Now, for any $g, h \in G,$ $gh = (gh)^{-1} = h^{-1} g^{-1} = hg.$ 4. Let $g_1, g_2 \in G.$ Then $|g_1 g_2| = |g_2 g_1|.$ ***Sol.*** First, observe that for any $x, y \in G,$ $xy = 1 \iff yx = 1.$ Now, for any $n \in \mathbb N,$ $(g_1 g_2)^n = g_1 (g_2 g_1)^{n-1} g_2.$ Hence, $(g_1 g_2)^n = 1$ $\iff$ $g_1 (g_2 g_1)^{n-1} g_2 = 1$ $\iff$ $(g_2 g_1)^{n-1} g_2 g_1 = 1$ $\iff$ $(g_2 g_1)^n = 1.$ 5. (i) Let $g \in G$ with $|g| = n < \infty.$ Then for every integer $m,$ $|g^m| = n / \gcd(m, n).$ (ii) If $G$ is a cyclic group of finite order $n$ then the number of distinct elements which generate $G$ is $\varphi(n),$ where $\varphi$ is Euler's totient function: that is, $\varphi(n)$ is the number of positive integers not exceeding $n$ which are co-prime to $n.$ ***Sol.*** (i) Let $k = n / \gcd(m, n).$ Since $m / \gcd(m, n)$ is an integer, and $g^n = 1,$ it follows that $mk \mid n,$ and hence $(g^m)^k = 1.$ On the other hand, suppose that $(g^m)^l = 1,$ for some integer $l.$ Then $|g| = n \implies n \mid ml \implies k \mid ml / \gcd(m, n) \implies k \le l.$ Thus, $k$ is the least positive integer such that $(g^m)^k = 1.$ (ii) Let $G = \langle g \rangle,$ where $|g| = n.$ For any element $g^m \in G,$ $G = \langle g^m \rangle$ $\iff$ $|g^m| = n$ $\iff$ $n / \gcd(m,n) = n$ $\iff$ $\gcd(m, n) = 1.$ 6. Let $g_1, g_2 \in G$ with $|g_1| = n_1 < \infty,$ $|g_2| = n_2 , \infty.$ If $n_1$ and $n_2$ are co-prime and $g_1$ and $g_2$ commute, then $|g_1 g_2| = n_1 n_2.$ ***Sol.*** First, $(g_1 g_2)^{n_1 n_2} = (g_1^{n_1})^{n_2} (g_2^{n_2})^{n_1} = 1.$ On the other hand, suppose $m$ is a positive integer such that $(g_1 g_2)^m = 1.$ Then $g_1^m = (g_2^m)^{-1}.$ Hence, $|g_1^m| = |g_2^m|,$ which implies that $n_1 / \gcd(n_1, m) = n_2 / \gcd(n_2, m).$ Since $n_1$ and $n_2$ are co-prime, this is possible only if $\gcd(n_1, m) = n_1$ and $\gcd(n_2, m) = n_2,$ which implies that $n_1 \mid m$ and $n_2 \mid m.$ Therefore, $n_1 n_2 = \lcm(n_1, n_2) \mid m$ . Thus, $n_1 n_2$ is the least positive integer $m$ such that $(g_1 g_2)^m = 1\text.$ 7. Let $g \in G$ with $|g| = n_1 n_2,$ where $n_1$ and $n_2$ are co-prime positive integers. Then there are elements $g_1, g_2 \in G$ such that $g = g_1 g_2 = g_2 g_1$ and $|g_1| = n_1,$ $|g_2| = n_2.$ Moreover, $g_1$ and $g_2$ are uniquely determined by these conditions. ***Sol.*** Since $n_1$ and $n_2$ are co-prime, there exist integers $a$ and $b$ such that $a n_1 + b n_2 = 1.$ Let $g_1 = g^{b n_2}$ and $g_2 = g^{a n_1}.$ Then, \begin{align*} g & = g^{a n_1 + b n_2} \\ & = g^{a n_1} g^{b n_2} \\ & = g_2 g_1 \\ & = g_1 g_2, \end{align*} since any two powers of $g$ commute. Also, $a n_1 + b n_2 = 1$ implies that $n_1$ and $b$ are co-prime. Therefore, $|g_1| = n_1 n_2/ \gcd(b n_2, n_1 n_2) = n_1 n_2 / n_2 = n_1.$ Similarly, $|g_2| = n_2.$ Now, suppose $h_1, h_2 \in G$ satisfy $g = h_1 h_2 = h_2 h_1$ and $|h_1| = n_1,$ $|h_2| = n_2.$ Then $g^{b n_2} = (h_1 h_2)^{b n_2} = h_1^{b n_2} = h_1^{1 - a n_1} = h_1.$ Similarly, $g^{a n_1} = h_2.$ Hence, $g_1$ and $g_2$ are uniquely determined by the given conditions. 8. By considering orders of elements in the multiplicative group of all non-zero elements of the field $\mathbb Z_p,$ prove Fermat's theorem: for every integer $a$ not divisible by $p,$ $a^{p-1} \equiv 1 \mod p.$ ***Sol.*** The multiplicative group of non-zero elements of $\mathbb Z_p$ has order $p - 1.$ If $a$ is not divisible by $p,$ then $a$ and $p$ are co-prime, and hence $a \equiv b \mod p,$ where $b \in \mathbb Z_p$ with $b$ and $p$ co-prime. Hence, $b$ is in the mutliplicative group, which implies that $b^{p-1} = 1.$ Therefore, $a^{p-1} \equiv 1 \mod p.$ 9. Let $G$ be an Abelian group of finite order $n.$ Show that the product of the $n$ distinct elements of $G$ is equal to the product of all elements of $G$ of order $2$ (where the latter product is interpreted as $1$ if $G$ has no element of order $2$). By applying this result to the multiplicative group of all non-zero elements of the field $\mathbb Z_p,$ prove Wilson's theorem for prime numbers: $(p - 1)! \equiv -1 \mod p.$ ***Sol.*** Let $G = \{g_1, \ldots, g_n \}.$ For each $g_i \in G,$ $g_i^{-1} = g_j$ with $j \ne i$ if and only if $g_i^2 \ne 1.$ Therefore, in the product $g_1 g_2 \cdots g_n,$ every non-identity element of order other than $2$ cancels out with its inverse. Hence, this product is equal to the product of the elements of order $2.$ Now, let $G$ be the multiplicative group of the non-zero elements of $\mathbb Z_p,$ namely $\{1, \ldots, p - 1\}.$ Then $(p - 1)!$ is the product of all elements of $G.$ If $n \in G$ has order $2,$ then there exists a positive integer $k < p$ such that $n^2 = kp + 1 \implies kp = n^2 - 1 = (n + 1)(n - 1).$ But $n - 1$ and $k$ are co-prime to $p,$ and therefore $n + 1$ must be equal to $p.$ Hence, $n = p - 1$ is the only element of $G$ having order $2.$ Thus, $(p - 1)! \equiv -1 \mod p.$ 10. Let $H \le G$ and let $g_1, g_2 \in G.$ Then $Hg_1 = Hg_2$ if and only if $g_1^{-1}H = g_2^{-1}H.$ ***Sol.*** Recall that $Hx = Hy$ is equivalent to $xy^{-1} \in H,$ and that $xH = yH$ is equivalent to $x^{-1} y \in H.$ Now, $Hg_1 = Hg_2 \iff g_1 g_2^{-1} \in H,$ that is, $\left(g_1^{-1}\right)^{-1}\left(g_2^{-1}\right) \in H,$ which is equivalent to $g_1^{-1}H = g_2^{-1}H.$ 11. (i) Let $J \le H \le G.$ Then $|G : J|$ is finite if and only if $|G : H|$ and $|H : J|$ are both finite, and if so, $|G : J| = |G : H| |H : J|.$ (ii) Let $J \le H \le G$ with $|G : J| = p,$ prime. Then either $H = J$ or $H = G.$ ***Sol.*** (i) Let $\{\, x_i \}_{i \in I}$ be the distinct representatives of left cosets of $H$ in $G,$ and let $\{\, y_r \,\}_{r \in R}$ be the disinct representatives of $J$ in $H.$ Observe that if $y_r J \ne y_s J,$ then $x_i y_r J \ne x_i y_s J,$ since $(x_i y_r)^{-1} (x_i y_s) = y_r^{-1} y_s \notin J.$ Also, if $x_i H$ and $x_j H$ are distinct, then they are disjoint, which implies that $x_i y_r J \ne x_j y_s J,$ for any $r, s \in R.$ Thus, $\{\, x_i y_r \,\}_{i \in I, r \in R}$ are distinct representatives of left cosets of $J$ in $G.$ It remains to show that these are form an exhaustive system of distinct representatives of left cosets of $J$ in $G.$ But this is obvious, since, for any $g \in G,$ $g = x_i a$ for some $i \in I,$ and $a \in H,$ and then $h = y_r b$ for some $b \in J,$ so $g = x_i y_r b \in x_i y_r J.$ Now, since$\{\, x_i y_r \,\}_{i \in I, r \in R} \cong \{\, x_i \,\}_{i \in I} \times \{\, y_r \,\}_{r \in R}$ (as sets), it follows that $|G : J| = |G : H| |H : J|,$ in terms of cardinalities, and in particular, the LHS is finite if and only if both the terms on the RHS are finite. (ii) Since $p = |G : J| = |G : H| |H : J|,$ either $|G : H| = 1$ or $|H : J| = 1.$ In the former case, $H = G,$ and in the latter, $J = H.$ 12. Let $H \le G$ and $g \in G.$ If $|g| = n$ and $g^m \in H,$ where $n$ and $m$ are co-prime integers, then $g \in H.$ ***Sol.*** There exist integers $a$ and $b$ such that $am + bn = 1.$ Hence, $g^{am} = g^{1 - bn} = g \in H$ (since $g^m \in H$). ## Chapter 2 13. Give an example of a semigroup without an identity element. ***Sol.*** The set $\mathbb N = \{1, 2, \ldots\}$ of natural numbers, under addition. 14. Give an example of an infinite semigroup with an identity element $e$ such that no element except $e$ has an inverse. ***Sol.*** The set $\mathbb N$ under multiplication. 15. Let $S$ be a semigroup and let $x \in S.$ Show that $\{x\}$ forms a subgroup of $S$ (of order $1$) if and only if $x^2 = x.$ Such an element $x$ is called an *idempotent* in $S.$ (Warning. A semigroup may have several different subgroups of order $1$: see <a href="#Q16">16</a>. Why does a group have only one subgroup of order $1$?) ***Sol.*** If $G = \{x\}$ is a subgroup of $S,$ then $x^2 \in G \implies x^2 = x.$ Conversely, if $x^2 = x,$ then $x$ is an identity element in $G.$ Since this is unique element of $G,$ $G$ is a group, and hence is a subgroup of $S.$ A group has only one subgroup of order $1,$ which is $\{e\},$ since for any element $x$ of a group, $x^2 = x \implies x = e,$ by cancellation. 16. Let $X$ be any non-empty set. Let $S$ be the set of all subsets of $X.$ Show that $S$ is a semigroup with respect to the operation $\cap.$ Does $S$ have an identity element, and if so, what are the units in $S$? Show that every element of $S$ is an idempotent (<a href="#Q15">15</a>). Deduce that for all $Y \in S,$ $\{Y\}$ is a subgroup of $S,$ and that every subgroup of $S$ has order $1.$ What happens if $\cap$ is replaced by $\cup$? ***Sol.*** The intersection of any two subsets of $X$ is a subset of $X,$ and $\cap$ is associative, hence $\cap$ is an associative binary operation on $S,$ which shows that $(S, \cap)$ is a semigroup. For any $Y \in S,$ $Y \subseteq X \implies Y \cap X = Y.$ Hence, $X$ (which is a subset of itself, and is therefore in $S$) is the identity element of $S.$ If $Y \subsetneq X,$ then $Y \cap Z \subsetneq X$ for all $Z \subseteq X,$ which shows that $Y$ cannot be a unit. Thus, $X$ is the only unit of $S.$ The intersection of any set with itself is itself, hence every element of $S$ is idempotent. Therefore, for any $Y \in S,$ $Y^2 = Y \cap Y = Y$ implies that $\{ Y \}$ is a subgroup of $S.$ A group can only have one idempotent elements, but since all elements of $S$ are idempotent, no two of them can be in the same subgroup of $S.$ Thus, all subgroups of $S$ have order $1.$ If $\cap$ is replaced by $\cup,$ all the above results hold, except that the identity element becomes $\varnothing$ instead of $X.$ 17. Let $X$ be a set with $|X| = 2.$ Do two elements of $M_X$ necessarily commute? Do two elements of $\Sigma_X$ necessarily commute? **Note.** $M_X$ is the semigroup of all endofunctions of $X,$ and $\Sigma_X$ is its group of units. ***Sol.*** No, not all elements of $M_X$ mutually commute. Let $X = \{1, 2\},$ let $f$ be the endofunction that maps both elements of $1,$ and let $g$ be the endofunction that maps both elements to $2.$ Then $fg = f \ne g = gf.$ $\Sigma_X = \{(), (12)\},$ and therefore any two elements of $\Sigma_X$ commute (since, e.g., every element commutes with itself and with the identity). 18. Express the permutation $\sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 5 & 7 & 4 & 6 & 2 & 3\end{pmatrix}$ as a product of disjoint cycles. Find expressions for $\sigma^2$ and $\sigma^{-1}$ as products of disjoint cycles. ***Sol.*** $\sigma = (256)(37),$ $\sigma^2 = (265),$ and $\sigma^{-1} = (265)(37).$ 19. If $\varphi \colon G \to H$ and $\psi \colon H \to J$ are homomorphisms, then the composite map $\varphi \psi \colon G \to J$ is also a homomorphism. ***Sol.*** For all $x, y \in G,$ \begin{equation*} (xy)(\varphi\psi) = ((xy)\varphi)\psi = ((x \varphi)(y \varphi))\psi = ((x \varphi) \psi)((y \varphi) \psi) = x(\varphi\psi) y(\varphi\psi). \end{equation*} 20. Verify that if $\mathcal G$ is any set of groups then $\cong$ is an equivalence relation on $\mathcal G.$ ***Sol.*** The identity map of any group $G$ is an isomorphism from $G$ to itself, hence $G \cong G.$ If $G \cong H,$ then there is an isomorphism $\varphi \colon G \to H,$ and an inverse isomorphism $\psi \colon H \to G,$ which implies that $H \cong G.$ Finally, if $G \cong H$ and $H \cong J,$ then there exist isomorphisms $\varphi \colon G \to H$ and $\psi \colon H \to J.$ Then $\varphi \psi$ is a homomorphism from $G$ to $J,$ and since the composition of bijections is a bijection, $\varphi\psi \colon G \to J$ is an isomorphism. Hence, $G \cong J.$ 21. (i) Calculate the products $(12)(13)(14)$ and $(12)(13)(14)(15)$ in the group $\Sigma_n,$ where $n$ is an integer with $n \ge 5.$ (ii) A permutation such as $(12),$ which interchanges two points and fixes all others, is called a *transposition*. For any integer $n > 1,$ show that the permutation $(123 \cdots n)$ can be expressed as a product of $n - 1$ transpositions. (iii) For any integer $n > 1,$ prove that every non-trivial element of $\Sigma_n$ can be expressed as a product of at most $n - 1$ transpositions. ***Sol.*** (i) $(12)(13)(14) = (1234),$ and $(12)(13)(14)(15) = (1234)(15) = (12345).$ (ii) $(123 \cdots n) = (12)(13) \cdots (1n),$ by induction on $n.$ (iii) We know that any non-trivial element of $\Sigma_n$ can be expressed as a sum of disjoint cycles, such that the sum of the lengths of the cycles (including cycles of length $1$) is $n.$ From (ii), any cycle of length $k$ can be expressed as a product of $k - 1$ transpositions. Thus, any non-trivial element of $\Sigma_n$ can be expressed as product of $n - c \le n - 1$ transpositions, where $c$ is the number of cycles in its cycle decomposition. 22. Let $n$ be a positive integer and let $\sigma \in \Sigma_n.$ Suppose that $\sigma$ can be expressed as a product of $s$ disjoint cycles of lengths $n_1, n_2, \ldots, n_s$ respectively, where $s, n_1, \ldots, n_s$ are positive integers such that $n_1 + \cdots + n_s = n.$ Then $|\sigma|$ is the least common multiple of $n_1, n_2, \ldots, n_s.$ ***Sol.*** If a permutation is a cycle of length $k,$ then its order is $k.$ Let $\sigma_1, \ldots, \sigma_s$ be the cycles in the cycle decomposition of $\sigma.$ Since disjoint cycles commute, $\sigma^m = \sigma_1^m \cdots \sigma_s^m.$ Moreover, if $\sigma_i$ and $\sigma_j$ are disjoint, then so are $\sigma_i^m$ and $\sigma_j^m,$ and hence, $\sigma^m = 1 \iff \sigma_i^m = 1$ for each $i = 1, \ldots, s.$ Thus, $m$ is a multiple of $n_i = |\sigma_i|$ for all $i = 1, \ldots, s.$ Then $|\sigma|,$ being the least such positive integer, is the least common multiple of $n_1, \ldots, n_s.$ 23. Let $n$ and $m$ be positive integers such that $m$ divides $n.$ Let $\sigma$ be a cycle of length $n$ in $\Sigma_n.$ Then $\sigma^m$ is the product of $m$ disjoint cycles of length $n/m.$ ***Sol.*** Let $n = mk,$ and without loss of generality, let $\sigma = (12 \cdots n).$ Now, $\sigma^m(1) = m + 1,$ $\sigma^m(2) = 2m + 1,$ $\ldots,$ $\sigma^m((k-1)m + 1) = 1.$ Hence, $(1, m + 1, \ldots, (k-1)m + 1)$ $= \sigma_1$ (say) is a cycle of $\sigma^m.$ Next, $\sigma^m(2) = m + 2,$ $\ldots,$ $\sigma^m((k - 1)m + 2) = 2,$ so $(2, m + 2, \ldots, (k-1)m + 2) = \sigma_2$ is a cycle of $\sigma^m.$ Proceeding similarly, $(m, 2m, \ldots, km) = \sigma_m$ is a cycle of $\sigma^m,$ and $km = n.$ Thus the cycle decomposition of $\sigma^m$ is $\sigma_1 \sigma_2 \cdots \sigma_m,$ with each cycle $\sigma_i$ having length $k = n/m.$ 24. Any group which can be embedded in an Abelian group is Abelian. ***Sol.*** Let $\varphi \colon G \to H$ be an embedding of $G$ into an Abelian group $H.$ Then for any $x, y \in G,$ $(xy)\varphi = (x\varphi)(y\varphi) = (y\varphi)(x\varphi) = (yx)\varphi,$ and since $\varphi$ is injective, this implies that $xy = yx.$ Therefore, $G$ is Abelian. 25. Prove that if $X$ is a non-empty set and $Y$ a non-empty subset of $X$ then $\Sigma_Y$ can be embedded in $\Sigma_X.$ (In particular, whenever $m$ and $n$ are positive integers such that $m \le n$ then $\Sigma_m$ can be embedded in $\Sigma_n.$) Hence or otherwise show that if $|X| > 2$ then $\Sigma_X$ is non-Abelian. ***Sol.*** Define a homomorphism $\varphi \colon \Sigma_Y \to \Sigma_X$ as follows: For each $\sigma \in \Sigma_Y,$ let $\sigma \varphi$ be the permutation of $X$ defined by \begin{align*} (x)(\sigma \varphi) = \begin{cases} x\sigma, & x \in Y \\ x, & x \notin Y. \end{cases} \end{align*} To see that $\varphi$ is a homomorphism, first note that $x(\sigma \varphi) \in Y \iff x \in Y.$ Now, let $\sigma_1, \sigma_2 \in \Sigma_Y,$ and let $x \in X.$ If $x \in Y,$ then $x((\sigma_1 \sigma_2)\varphi) = x(\sigma_1 \sigma_2) = (x\sigma_1)\sigma_2 = (x(\sigma_1 \varphi))(\sigma_2 \varphi).$ If $x \notin Y,$ then $x(\sigma_1 \varphi) = x \implies (x(\sigma_1 \varphi))(\sigma_2 \varphi) = x = x((\sigma_1 \sigma_2)\varphi).$ Thus, $(\sigma_1 \sigma_2)\varphi = (\sigma_1 \varphi)(\sigma_2 \varphi).$ Moreover, $\varphi$ is injective: for any $\sigma_1, \sigma_2 \in \Sigma_Y,$ and $y \in Y,$ if $y(\sigma_1\varphi) = y \sigma_1 = y \sigma_2 = y(\sigma_2\varphi),$ then $\sigma_1 = \sigma_2.$ Therefore, $\varphi$ is an embedding of $\Sigma_Y$ in $\Sigma_X.$ Now, if $|X| > 2,$ then let $Y \subseteq X$ have three elements. Then $\Sigma_Y \cong \Sigma_3$ is non-Abelian, and since $\Sigma_Y$ embeds in $\Sigma_X,$ the latter is non-Abelian too (see <a href="#Q24">24</a>). 26. If $\varphi \colon G \to H$ is a homomorphism and $G$ is Abelian then $\Im \varphi$ is Abelian. ***Sol.*** Let $x\varphi, y\varphi \in \Im \varphi.$ Then $(x\varphi)(y\varphi) = (xy)\varphi = (yx)\varphi = (y\varphi)(x\varphi).$ 27. Suppose that $G_1$ and $G_2$ are isomorphic groups, and let $\varphi \colon G_1 \to G_2$ be an isomorphism. If $K_1 \le G_1$ and $K_2 = K_1 \varphi$ then $|G_1 : K_1| = |G_2 : K_2|.$ ***Sol.*** Let $\{x_i\}_{i \in I}$ be a system of distinct representatives of left cosets of $K_1$ in $G_1.$ For each left coset $x_i K_1,$ $(x_i K_1)\varphi = (x_i \varphi)K_2$ is a left coset of $K_2$ in $G_2.$ If $(x_i \varphi)K_2 = (x_j \varphi)K_2,$ then $(x_i^{-1} x_j)\varphi = (x_i \varphi)^{-1}(x_j \varphi) \in K_2 = K_1 \varphi,$ and therefore $x^{-1} x_j \in K_1$ (since $\varphi$ is injective), which implies that $x_i K_1 = x_j K_2.$ Thus, $\varphi$ maps distinct left cosets of $K_1$ to distinct left cosets of $K_2.$ Since $\varphi$ is surjective, any element of any coset of $K_2$ can be written as $x \varphi,$ for some $x \in G_1,$ and then $x \in x_i K_1,$ for some $i \in I.$ Thus, $x \varphi \in (x_i \varphi) K_2,$ which shows that every left coset of $K_2$ in $G_2$ is of the form $(x_i \varphi) K_2,$ for some $i \in I.$ Hence, the image of $\{x_i\}_{i \in I}$ under $\varphi$ is a system of distinct representatives of left cosets of $K_2$ in $G_2,$ and is in bijection with $\{x_i\}_{i \in I}.$ Therefore, $|G_1 : K_1| = |G_2 : K_2|.$ 28. If $G$ and $H$ are finite, then a necessary condition that $G$ can be embedded in $H$ is that $|G|$ divides $|H|.$ Show by example that this is not a sufficient condition. ***Sol.*** We know that if $G$ can be embedded in $H,$ then $G$ is isomorphic to a subgroup of $H.$ Thus, $|G|$ must be the order of a subgroup of $H,$ which implies that $|G|$ divides $|H|.$ To see that this condition is not sufficient, let $G = \Sigma_3,$ which is of order $6,$ and let $H$ be the cyclic group of order $6.$ Then $|G| = |H|,$ but $G,$ which is non-Abelian, cannot be embedded in $H,$ which is Abelian (see <a href="#Q24">24</a>). 29. (i) Any infinite cyclic group is isomorphic to $\mathbb Z^+.$ (ii) If $G$ is a non-trivial group of which the only subgroups are $1$ and $G$ then $G$ is cyclic of prime order. ***Sol.*** Let $G = \langle g \rangle$ be an infinite cyclic group. Define $\varphi \colon G \to \mathbb Z^+$ by $(g^n)\varphi = n,$ for each $g^n \in G.$ Note that $g^n = g^m$ if and only if $g^{n - m} = 1,$ which is equivalent to $n = m,$ since $g$ has infinite order. Thus, $\varphi$ is well-defined and injective. Finally, for any $n \in \mathbb Z^+,$ $(g^n)\varphi = n,$ hence $\varphi$ is surjective. Thus, $\varphi$ is an isomorphism from $G$ to $\mathbb Z^+.$ (ii) Let $g \in G$ be non-trivial. Then $\langle g \rangle,$ being a non-trivial subgroup of $G,$ must be equal to $G.$ Now, let $|G| = |g| = mn,$ by any factorisation of $|G|$ into two positive integers. Then $\langle g^m \rangle$ is a subgroup of $G$ having order $n,$ which implies that $n = 1$ or $|G|.$ That is, the only divisors of $|G|$ are $1$ and itself, which shows that $|G|$ ($\ne 1$) is prime. 30. (i) If $G$ is finite then no proper subgroup of $G$ is isomorphic to $G.$ (ii) If $G$ is a group such that whenever $J < H \le G,$ $J \not\cong H$ then every element of $G$ has finite order. (An example of an infinite group with this property will be given in <a href="#Q144">144</a>.) (i) Isomorphic groups have the same order, but a proper subgroup of a finite group has order strictly less than the order of the group. (ii) Suppose, to the contrary, that there exists an element $g \in G$ having infinite order. Let $H = \langle g \rangle$ and $J = \langle g^2 \rangle.$ Then $J < H,$ but $J \cong \mathbb Z^+ \cong H,$ which contradicts the given condition. 31. Show that $\mathbb Z^+$ has infinitely many distinct subgroups. Deduce that every infinite group has infinitely many distinct subgroups. ***Sol.*** For each $n \in \mathbb N,$ we have a subgroup $\langle n \rangle \le \mathbb Z^+.$ For any integer $m,$ $m \in \langle n \rangle$ if and only if $m = nk,$ for some integer $k \in \mathbb Z,$ i.e., $n \mid m.$ Thus, $\langle \mathbb n \rangle = \langle \mathbb m \rangle$ only if $m \mid n$ and $n \mid m,$ which implies that $m = n$ (since $m, n \in \mathbb N$). Thus, $\{\, \langle n \rangle \mid n \in \mathbb N \,\}$ is a collection of infinitely many distinct subgroups of $\mathbb Z^+.$ Now, let $G$ be any infinite group. If $G$ contains an element, say $g,$ of infinite order, then $\langle g \rangle \cong \mathbb Z^+$ contains infinitely many distinct subgroups, all of which are subgroups of $G.$ On the other hand, suppose every element of $G$ has finite order. Then consider the collection of cyclic subgroups generated by all the elements of $G.$ If $x \in G,$ then, since $\langle x \rangle$ is finite, there are infinitely many $y \in G$ such that $\langle x \rangle \ne \langle y \rangle.$ Thus, $G$ has infinitely many distinct subgroups. 32. No two of the groups $\mathbb Z^+,$ $\mathbb Q^+,$ $\mathbb R^+$ are isomorphic. (Remark. It is in fact true that $\mathbb R^+ \cong \mathbb C^+.$ This can be proved by regarding $\mathbb R$ and $\mathbb C$ as vector spaces over $\mathbb Q,$ and using vector space theory and facts about infinite cardinals.) ***Sol.*** Let $\varphi \colon \mathbb Z^+ \to \mathbb Q^+$ be any injective homomorphism, and let $r = 1\varphi \in \mathbb Q^+.$ Then, $r/2 \in \mathbb Q^+,$ but there exists no $n \in \mathbb Z^+$ such that $r/2 = n\varphi,$ since otherwise $1\varphi = r = 2(r/2) = 2(n\varphi) = (2n)\varphi$ would imply that $2n = 1$ (by injectivity of $\varphi$), but $\mathbb Z^+$ contains no such $n.$ Therefore, $\varphi$ is not surjective. This shows that there is no isomorphism from $\mathbb Z^+$ to $\mathbb Q^+.$ Since $\mathbb Q^+ \le \mathbb R^+,$ it follows that $\mathbb Z^+$ cannot be isomorphic to $\mathbb R^+$ either. Next, suppose that $\psi \colon \mathbb Q^+ \to \mathbb R^+$ is an isomorphism. For any $m/n \in \mathbb Q^+,$ $(m/n)\psi = y \in \mathbb R^+$ implies that $m \psi = n ((m/n)\psi) = ny.$ It follows that if $1\psi = x \in \mathbb R^+,$ then $(m/n)\psi = mx/n \in \mathbb R^+.$ Now, let $m/n \in \mathbb Q^+$ be such that $\sqrt 2 = (m/n)\psi = mx/n,$ and let $p/q \in \mathbb Q^+$ be such that $2 = (p/q)\psi = px/n.$ Then $\sqrt 2 = 2 / \sqrt 2 = (px/n)/(mx/n) = (pn)/(mq),$ a rational number, which is a contradiction. **Note.** This proof is given to demonstrate that a purely algebraic proof is possible. The simplest argument, of course, is that $\mathbb Q^+$ is countable and $\mathbb R^+$ is uncountable, and hence there is no bijection between them, let alone an isomorphism of groups. 33. Let $\Hom(G, A)$ denote the set of all homomorphisms of a group $G$ into an Abelian group $A.$ Define a binary operation $+$ on $\Hom(G, A)$ as follows: for $\varphi, \psi \in \Hom(G, A),$ \begin{equation*} \varphi + \psi \colon G \to A \end{equation*} is defined by \begin{equation*} \varphi + \psi \colon g \mapsto g^\varphi g^\psi \end{equation*} for all $g \in G.$ Verify that $\varphi + \psi \in \Hom(G, A)$ and that, with respect to $+,$ $\Hom(G, A)$ acquires the structure of an Abelian group. Show that for any Abelian group $A,$ $\Hom(\mathbb Z^+, A) \cong A.$ ***Sol.*** Let $\varphi, \psi \in \Hom(G, A).$ Then, for any two elements $g, h \in G,$ \begin{align*} (gh)(\varphi + \psi) & = (gh)^\varphi (gh)^\psi \\ & = g^\varphi h^\varphi g^\psi h^\psi \\ & = g^\varphi g^\psi h^\varphi h^\psi \qquad \text{[$A$ is Abelian]} \\ & = g(\varphi + \psi) h(\varphi + \psi). \end{align*} Thus, $\varphi + \psi \in \Hom(G, A).$ Associativity and commutativity of $+$ in $\Hom(G, A)$ follow from associativity and commutativity of multiplication in $A.$ The trivial homomorphism from $G$ to $A$ is the clearly identity element of $\Hom(G, A).$ For any $\varphi \in \Hom(G, A),$ the map $\psi \colon G \to A$ defined by $g\psi = (g\varphi)^{-1},$ for all $g \in G,$ is a homomorphism from $G$ to $A$ and is an inverse with respect to $+,$ as shown below: \begin{align*} g(\varphi + \psi) & = g^{\varphi} g^{\psi} \\ & = g^{\varphi} (g^\varphi)^{-1} \\ & = 1 \end{align*} which implies that $\varphi + \psi$ is the trivial homomorphism from $G$ to $A,$ the identity element of $\Hom(G, A).$ Therefore, $\Hom(G, A)$ is an Abelian group with respect to $+.$ Now, for each $a \in A,$ let $\varphi_a \colon \mathbb Z^+ \to A$ be the function defined by $n\varphi = a^n,$ for all $n \in \mathbb Z^+.$ Then $\varphi_a$ is a homomorphism from $\mathbb Z^+$ to $A,$ since \begin{equation*} (m + n)\varphi = a^{m + n} = a^m a^n = (m\varphi)(n\varphi). \end{equation*} Next, define a map $f \colon A \to \Hom(\mathbb Z^+, A)$ by $af = \varphi_a,$ for all $a \in A.$ First, observe that $f$ is a homomorphism: if $a, b \in A,$ $(ab)f = \varphi_{ab},$ and $af + bf = \varphi_a + \varphi_b.$ To see that these two maps are equal, let $n \in \mathbb Z^+.$ Then \begin{equation*} n \varphi_{ab} = (ab)^n = a^n b^n = n^{\varphi_a} n^{\varphi_b} = n(\varphi_a + \varphi_b). \end{equation*} Next, to show that $f$ is bijective, observe that every element $\varphi$ of $\Hom(\mathbb Z^+, A)$ is uniquely determined by the value of $1\varphi$: if $\varphi \in \Hom(\mathbb Z^+, A),$ then $n\varphi = n(1\varphi),$ which shows that every $\varphi \in \Hom(\mathbb Z^+, A)$ is of the form $\varphi_a,$ with $a = 1\varphi,$ and that $\varphi_a = \varphi_b \iff a = b.$ Therefore, $f$ is an isomorphism from $A$ to $\Hom(\mathbb Z^+, A).$ 34. Show that if $\varphi \colon G \to H$ is a homomorphism, $g \in G,$ and $n$ is a positive integer such that $g^n = 1,$ then $(g\varphi)^n = 1.$ Hence, by considering the elements $z$ of $\mathbb C^\times$ satisfying $z^3 = 1,$ or otherwise, show that $\mathbb C^\times$ is not isomorphic to either $\mathbb Q^\times$ or $\mathbb R^\times.$ ***Sol.*** $g^n = 1 \implies (g^n)\varphi = 1\varphi \implies (g\varphi)^n = 1.$ Now, let $\varphi \colon \mathbb C^\times \to G$ be any homomorphism, where $G = \mathbb Q^\times$ or $\mathbb R^\times.$ Let $z = e^{2\pi i / 3} \in \mathbb C^\times.$ Then $z^3 = 1 \implies (z\varphi)^3 = 1.$ But the only such element in $G$ is $1,$ hence $z\varphi = 1 = 1\varphi,$ which shows that $\varphi$ is not injective, and hence cannot be an isomorphism. 35. Prove that $\mathbb Q^\times$ is not isomorphic to $\mathbb R^\times.$ (Hint. Note that for any positive real number $a$ and any positive integer $n,$ there is a unique positive real number $b$ such that $a = b^n.$) ***Sol.*** Suppose, to the contrary, that there exists an isomorphism $\varphi \colon \mathbb R^\times \to \mathbb Q^\times.$ Then, $(-1)^2 = 1$ $\implies$ $((-1)\varphi)^2 = 1\varphi = 1$ $\implies$ $(-1)\varphi = -1,$ since $1$ and $-1$ are the only square roots of $1$ in $\mathbb Q,$ and $\varphi$ is injective. This implies that $(-x)\varphi = -(x \varphi)$ for all $x \in \mathbb R^\times.$ Now, let $x = 2 \varphi^{-1} \in \mathbb R^\times.$ If $x > 0,$ then $x = (\sqrt x)^2$ $\implies$ $2 = x\varphi = ((\sqrt x)\varphi)^2,$ which is impossible, as $\sqrt 2 \notin \mathbb Q.$ If $x < 0,$ then $-x = (\sqrt{-x})^2$ $\implies$ $-2 = (-x)\varphi = ((\sqrt{-x})\varphi)^2,$ which is impossible, as $\sqrt{-2} \notin \mathbb Q.$ Thus, there is no such isomorphism $\varphi.$ 36. Prove that there is no field $F$ such that $F^+ \cong F^\times.$ (Hint. Assume that there is a field $F$ with an isomorphism $\varphi \colon F^\times \to F^+$ and consider $(-1)\varphi.$) ***Sol.*** If possible, let $\varphi \colon F^\times \to F^+$ be an isomorphism, and let $v = (-1)\varphi \in F^+.$ Then, $(-1)^2 = 1$ $\implies$ $2v = 2((-1)\varphi) = 1\varphi = 0.$ This implies that $v = 0$ (for, if $v \ne 0,$ then $2v = 0$ $\implies$ $2 \times 1 = 0$ $\implies$ $1 = -1$ $\implies$ $v = 1\varphi = 0$), and therefore, $1 = -1.$ Now, for all $x \in F^\times,$ $(x^{-1})\varphi = -(x\varphi) = x\varphi$ $\implies$ $x^{-1} = x$ $\implies$ $x^2 = 1.$ But this implies $x^2 - 1 = 0$ $\implies$ $(x + 1)(x - 1) = 0$ $\implies$ $(x - 1)^2 = 0$ $\implies$ $x = 1.$ Hence, $F = \{0, 1\},$ and therefore $F^\times = \{1\}$ cannot be isomorphic to $F^+ = \{0, 1\}.$ 37. Let $G$ be a field. Define a binary operation $*$ on $F$ by \begin{equation*} a*b = a + b - ab \qquad \text{for all $a, b \in F.$} \end{equation*} Prove that the set of all elements of $F$ distinct from $1$ forms a group $F^*$ with respect to the operation $*,$ and that $F^* \cong F^\times.$ ***Sol.*** Let $F^* = \{\, a \in F \mid a \ne 1 \,\}.$ Observe that if $x, y \in F,$ then $(1 - x)(1 - y) = 1 - x - y + xy.$ Therefore, if $a, b \in F^*,$ then $x = 1 - a$ and $y = 1 - b$ are in $F^\times,$ and $a*b = 1 - xy.$ From this, it is easy to see that $*$ is an assocative and commutative binary operation on $F^*,$ and that $1 - 1 = 0$ is the identity element and for each $a \in F^*,$ $a*b = 0$ if $1 - (1 - a)(1 - b) = 0,$ i.e., $b = \frac a {a - 1}$ (which is clearly well-defined and not equal to $1$), so every element of $F^*$ has an inverse with respect to $*.$ Therefore, $F^*$ is a group. Now, define $\varphi \colon F^\times \to F^*$ by $x\varphi = 1 - x,$ for all $x \in F^\times.$ This is well defined, since $x \in F^\times$ $\implies$ $x \ne 0$ $\implies$ $1 - x \ne 1.$ It is also obvious that $\varphi$ is bijective, with inverse $a\varphi^{-1} = 1 - a,$ for all $a \in F^*.$ The earlier observation shows that $(xy)f = (xf)*(yf),$ hence $f$ is an isomorphism. 38. The set of all positive rational numbers forms a subgroup $\mathbb Q^+_{\text{pos}}$ of $\mathbb Q^\times,$ and there is a surjective homomorphism \begin{equation*} |~~| \colon \mathbb Q^\times \to \mathbb Q^\times_{\text{pos}} \end{equation*} which is not an isomorphism. Is $\mathbb Q^\times_{\text{pos}} \cong \mathbb Q^+$? ***Sol.*** For any two rational numbers $a$ and $b,$ $|ab| = |a||b|,$ which shows that $|~~|$ is a homomorphism. For each positive rational number $a,$ $a = |a|,$ hence $|~~|$ is a surjection from $\mathbb Q^\times$ to $\mathbb Q^\times_{\text{pos}}.$ Finally, $1, -1 \in \mathbb Q^\times,$ and $|-1| = |1| = 1,$ hence $|~~|$ is not an isomorphism. To see that $\mathbb Q^\times_{\text{pos}} \not\cong \mathbb Q^+,$ suppose, to the contrary, that there exists an isomorphism $\varphi \colon \mathbb Q^\times_{\text{pos}} \to \mathbb Q^+.$ Let $a = 2\varphi.$ Then $a/2 \in \mathbb Q^\times_{\text{pos}}$ $\implies$ $a/2 = x\varphi,$ for some $x \in \mathbb Q^\times_{\text{pos}}.$ But $2\varphi = a = 2(a/2) = 2(x\varphi) = (x^2)\varphi$ $\implies$ $x^2 = 2,$ which is impossible, since $\sqrt 2 \notin \mathbb Q^\times_{\text{pos}}.$ Hence, $\mathbb Q^\times_{\text{pos}} \not\cong \mathbb Q^+.$ 39. For any positive integer $n,$ the only homomorphism $\varphi \colon C_n \to \mathbb C^+$ is the trivial homomorphism. ***Sol.*** Let $\varphi \colon C_n \to C^\times$ be any homomorphism, and let $z = (e^{2i\pi/n})\varphi.$ Then, $(e^{2i\pi/n})^n = 1$ $\implies$ $nz = 1\varphi = 0$ $\implies$ $z = 0.$ Since every element of $C_n$ is a power of $e^{2i\pi/n},$ this implies that $\varphi$ maps all elements of $C_n$ to $0,$ and is therefore the trivial homomorphism. 40. Let $n$ be a positive integer. Then (i) $\mathbb Z_n^+ \cong C_n.$ (ii) $\mathbb Z_n^\times$ is an Abelian group of order $\varphi(n),$ where $\varphi$ is Euler's function (see <a href="#Q5">5</a>). (iii) By considering orders of elements in $\mathbb Z_n^\times,$ prove the Euler-Fermat theorem: \begin{equation*} m^{\varphi(n)} \equiv 1 \mod n, \end{equation*} whenever $m$ is an integer co-prime to $n.$ (This generalizes <a href="#Q8">8</a>). ***Sol.*** (i) Let $\omega = e^{2i\pi/n},$ and define $\varphi \colon \mathbb Z_n^+ \to C_n$ by $k\varphi = \omega^k,$ for all $k \in \mathbb Z_n^+.$ Then for $k, l \in \mathbb Z_n^+,$ $(k + l)\varphi = \omega^{k + l},$ since $\omega^t = \omega^s$ whenever $t \equiv s \mod n.$ Thus, $(k + n)\varphi = (\omega^k)(\omega^l) = (k\varphi)(l\varphi).$ Since $|\mathbb Z_n^+| = |C_n|$ and $\varphi$ is clearly surjective, $\varphi$ is an isomorphism. (ii) Multiplication is associative and commutative in $\mathbb Z_n^\times,$ $1 \in \mathbb Z_n^\times,$ and all elements of $\mathbb Z_n^\times$ are invertible, by definition. If $a, b \in \mathbb Z_n^\times,$ then $(ab)(b^{-1}a^{-1}) = 1,$ hence $ab \in \mathbb Z_n^\times.$ Therefore, $\mathbb Z_n^\times$ is an Abelian group. Now, let $m \in \mathbb Z_n.$ First, suppose that $m$ is not co-prime to $n,$ so $\gcd(m, n) = d \ne 1.$ Then $n/d$ is a non-zero element of $\mathbb Z_n,$ and $m(n/d) = n(m/d) = 0$ (since $d \mid m$). Thus, $m$ is a zero-divisor and hence is not a unit of $\mathbb Z_n.$ Next, suppose that $m$ is co-prime to $n.$ Then there exist integers $a$ and $b$ such that $am + bn = 1$ $\implies$ $am \equiv 1 \mod n.$ Thus, $a$ (modulo $n$) is a multiplicative inverse of $m,$ and $m$ is a unit of $\mathbb Z_n.$ Therefore, $\mathbb Z_n^\times$ consists of all the positive integers less than $n$ and co-prime to it. Hence, $|\mathbb Z_n^\times| = \varphi(n).$ (iii) From (ii), if $m$ is co-prime to $n,$ then $m \equiv a \mod n$ for some $a \in \mathbb Z_n^\times.$ Then $m^{\varphi(n)} \equiv a^{\varphi(n)} \equiv 1 \mod n,$ since $\varphi(n) = |\mathbb Z_n^\times|.$ 41. (i) Let $G$ be a group and $n$ a positive integer such that $g^n = 1$ for all $g \in G.$ Show that if $\varphi \colon G \to \mathbb C^\times$ is a homomorphism then $\Im \varphi \le C_n.$ Hence show that $\Hom(G, \mathbb C^\times) \cong \Hom(G, C_n)$ (cf. <a href="#Q33">33</a>). (ii) Deduce that if $G$ is a finite group then so is $\Hom(G, \mathbb C^\times).$ (Remark. If $G$ is a finite Abelian group then in fact $\Hom(G, \mathbb C^\times) \cong G,$ but we do not yet have the means of proving this. See <a href="#Q454">454</a>.) ***Sol.*** (i) For each $g \in G,$ $g^n = 1$ $\implies$ $(g\varphi)^n = 1$ $\implies$ $g\varphi \in C_n.$ Now, let $f \colon \Hom(G, C_n) \to \Hom(G, \mathbb C^\times)$ be the inclusion map, which is always an injective homomorphism. As proved above, $f$ is surjective as well. Hence, $\Hom(G, \mathbb C^\times) \cong \Hom(G, C_n).$ (ii) If $G$ is finite, of order $n,$ then $g^n = 1$ for all $g \in G.$ Hence, by (i), $\Hom(G, \mathbb C^\times) \cong \Hom(G, C_n),$ which is finite since $G$ and $C_n$ are both finite. 42. Let $G$ be a finite group such that $g^2 = 1$ for all $g \in G.$ Show that $G \cong V^+$ for some finite dimensional vector space $V$ over $\mathbb Z_2.$ Deduce that $|G| = 2^m$ for some integer $m \ge 0.$ (Hint. By <a href="#Q3">3</a>, $G$ is Abelian. Let $V$ consist of the same elements as $G,$ with the sum of 2 elements of $V$ equal to their product in $G,$ and scalar multiplication defined in the obvious way.) ***Sol.*** Let $V = G.$ Define $+ \colon V \times V \to V$ by $g + h = gh,$ for all $g, h \in V.$ Then $(V, +)$ is clearly an Abelian group. Define $\cdot \colon \mathbb Z_2 \times V \to V$ by $n \cdot g = g^n,$ for all $n \in \mathbb Z_2$ and $g \in V.$ Now, for $a, b \in \mathbb Z_2$ and $g, h \in V,$ $(a + b) \cdot (g + h) = (gh)^{a + b} = g^a h^a g^b h^b = a \cdot g + a \cdot h + b \cdot g + b \cdot h,$ $a \cdot (b \cdot g) = (g^b)^a = g^{ab} = ab \cdot g,$ and $1 \cdot g = g^1 = g.$ Thus, $(V, +, \cdot)$ is a vector space over $\mathbb Z_2.$ Let $\varphi \colon G \to V$ be the function defined by $g\varphi = g,$ for all $g \in G.$ From the definition of $+,$ it is evident that $(gh)\varphi = g\varphi + h\varphi,$ for all $g, h \in G.$ Since $V = G,$ $\varphi$ is obviously a bijection. Hence. $\varphi$ is an isomorphim from $G$ to $V^+.$ Now, $V$ is a finite dimensional vector space over $\mathbb Z_2$ (since it is finite), and therefore is isomorphic to $\mathbb Z_2^m,$ for some integer $m \ge 0.$ Hence, $|G| = |V| = |Z_2|^m = 2^m.$ 43. Let $F$ be a field and let $m, n$ be positive integers with $m \le n.$ Then (i) $\GL_m(F)$ can be embedded in $\GL_n(F),$ (ii) $\GL_n(F)$ is non-Abelian for all $n \ge 2.$ ***Sol.*** (i) Let $k = n - m.$ Define a map $\varphi \colon \GL_m(F) \to \GL_n(F)$ by \begin{equation*} A\varphi = \begin{pmatrix} A & 0_{m,k}\\ 0_{k,m} & I_k \end{pmatrix} \end{equation*} where $0_{m,k}$ and $0_{k,m}$ are zero matrices of orders $k \times m$ and $m \times k$ respectively, and $I_k$ is the identity matrix of order $k.$ From the laws of multiplication of partitioned matrices, it is clear that $\varphi$ is a homomorphism. Since equality of two $n \times n$ matrices implies equality of all corresponding pairs of submatrices, $\varphi$ is an embedding. (ii) Let $A = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix},$ and $B = A^T.$ Then $A, B \in \GL_2(F)$ and $AB \ne BA,$ which implies that $\GL_2(F)$ is non-Abelian. From (i), $\GL_2(F)$ can be embedded in $\GL_n(F)$ for all $n \ge 2,$ and therefore, $\GL_n(F)$ is non-Abelian as well (see <a href="#Q24">24</a>). 44. $\GL_2(\mathbb Z_2) \cong \Sigma_3.$ ***Sol.*** Consider the vector space $V = \mathbb Z_2^2$ with the standard basis $B = \left\{ u = \begin{bmatrix}1\\0\end{bmatrix}, v = \begin{bmatrix}0\\1\end{bmatrix} \right\}.$ Then $\GL_2(\mathbb Z_2) \cong \GL(V),$ with the matrices acting on the vectors of $V$ by left-multiplication. Let $X = \{u, v, u + v\},$ and note that any two distinct vectors of $X$ are linearly independent. Thus, any mapping of $u$ and $v$ to a pair of distinct vectors of $X$ defines a unique automorphism of $V,$ and conversely each automorphism defines such a mapping. But such a mapping is equivalent to a permutation of $X.$ Therefore, every automorphism of $V$ corresponds to a unique permutation of the set $X,$ which implies that $\GL(V) \cong \Sigma_X.$ This further implies that $\GL_2(\mathbb Z_2) \cong \Sigma_3.$ 45. Let $G$ be the set of all matrices of the form $\begin{pmatrix}a & b\\0 & c\end{pmatrix},$ where $a,$ $b,$ $c$ are real numbers such that $ac \ne 0.$ Prove that $G$ forms a subgroup of $\GL_2(\mathbb R),$ and that the set $H$ of all elements of $G$ in which $a = c = 1$ forms a subgroup of $G$ isomorphic to $\mathbb R^+.$ Find all elements in $G$ of order $2.$ Hence show that the product of two elements of order $2$ in $G$ can be an element of infinite order. ***Sol.*** The determinant of any matrix in $G$ is $ac \ne 0,$ hence $G \subseteq \GL_2(\mathbb R),$ and $G$ is clearly non-empty. Now, let $A = \begin{pmatrix}a_1 & b_1\\0 & c_1\end{pmatrix}$ and $B = \begin{pmatrix}a_2 & b_2\\0 & c_2\end{pmatrix}$ be any two elements of $G.$ Then \begin{align*} AB^{-1} &= \begin{pmatrix}a_1 & b_1\\0 & c_1\end{pmatrix} \times \dfrac{1}{a_2 c_2} \begin{pmatrix}c_2 & -b_2\\0 & a_2\end{pmatrix} \tag{45.1}\label{eq:45.1}\\ & = \dfrac{1}{a_2 c_2} \begin{pmatrix} a_1 c_2 & a_2 b_1 - a_1 b_2 \\ 0 & a_2 c_1 \end{pmatrix}, \end{align*} hence $AB^{-1} \in G.$ Therefore, $G \le \GL_2(\mathbb R).$ Let $H$ be the set of all the matrices in $G$ whose diagonal elements are both $1.$ From $\eqref{eq:45.1},$ if $A, B \in H,$ then $AB = A(B^{-1})^{-1} = \begin{pmatrix}1 & b_1 + b_2\\0 & 1\end{pmatrix}.$ Define $\varphi \colon \mathbb R^+ \to G$ as the function that maps each $b \in \mathbb R^+$ to the matrix in $H$ whose $(2,1)$-entry is $b.$ This is clearly an injection, and the preceding observation shows that it is a homomorphism. It is also evident that $\Im \varphi = H.$ Hence, $H \le G$ and $H \cong \mathbb R^+.$ To find all elements of $G$ of order $2,$ let $A = \begin{pmatrix}a & b\\0 & c\end{pmatrix}.$ Then $A^2 = \begin{pmatrix}a^2 & b(a + c) \\ 0 & c^2 \end{pmatrix}.$ Therefore, $A^2 = I$ if and only if $|a| = |c| = 1$ and either $b = 0$ or $c = -a.$ Let $A = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$ and $B = \begin{pmatrix}1 & 1\\0 & -1\end{pmatrix}.$ Then $A$ and $B$ are of order $2,$ but $AB = \begin{pmatrix} 1 & 1\\0 & 1\end{pmatrix}$ is a non-identity element of $H \cong \mathbb R^+,$ and hence has infinite order. 46. (i) If $R$ is a ring with a multiplicative identity $1$ then $R^\times$ can be embedded in $\Aut R^+.$ (Hint. See 2.11.) (ii) $\Aut \mathbb Z^+ \cong \mathbb Z^\times$ and $\Aut \mathbb Z_n^+ \cong \mathbb Z_n^\times$ for every positive integer $n.$ ***Sol.*** (i) For each $u \in R^\times,$ define a function $\alpha_u \colon R^+ \to R^+$ by $x \alpha_u = xu,$ for all $x \in R^+.$ Distributivity implies that $\alpha_u$ is an endomorphism of $R^+.$ Since $u$ is a unit, it has a multiplicative inverse $u^{-1} \in R^\times,$ and for all $x \in R^+,$ $(x\alpha_u)\alpha_{u^{-1}} = (xu)u^{-1} = x,$ which implies that $\alpha_u \alpha_{u^{-1}}$ is the identity automorphism. Since $u^{-1} \in R^+$ and $(u^{-1})^{-1} = u,$ this also implies that $\alpha_{u^{-1}} \alpha_u$ is the identity automorphism. Thus, $\alpha_u$ is an automorphism of $R^+.$ Now, define a map $\varphi \colon R^\times \to \Aut R^+$ by $u\varphi = \alpha_u,$ for all $u \in R^\times.$ This is well-defined, as shown above. If $u, v \in R^\times,$ then $x \alpha_u \alpha_v = xuv = x\alpha_{uv},$ for all $x \in R^+.$ That is, $(u\varphi)(v\varphi) = (uv)\varphi,$ and hence $\varphi$ is a group homomorphism. If $u\varphi = v\varphi,$ then $\alpha_u = \alpha_v$ $\implies$ $u = 1u = 1\alpha_u = 1\alpha_v = 1v = v.$ Hence, $\varphi$ is an embedding. (ii) Let $R$ be the ring $\mathbb Z$ or $\mathbb Z_n.$ Let $\varphi$ be the embedding of $R^\times$ into $\Aut R^+$ defined in (i). Let $\alpha$ be any automorphism of $R^+,$ and let $m = 1\alpha.$ Note that $m = 1\alpha \implies m \in R^\times.$ Since $R^+ = \langle 1 \rangle,$ every element of $n \in R^+$ can be written in the form $n1$, and hence $(n1)\alpha = n(1\alpha) = nm = n\alpha_m.$ But $\alpha_m = m\varphi.$ This shows that $\varphi$ is surjective, and hence bijective. Therefore, $\Aut R^+ \cong R^\times,$ for $R = \mathbb Z$ or $\mathbb Z_n.$ Note. To see that $m \in \mathbb R^\times,$ observe that, since $\alpha$ is an automorphism, $m(1\alpha^{-1}) = (m1)\alpha^{-1} = m\alpha^{-1} = 1,$ which shows that $m^{-1} = 1\alpha^{-1},$ and hence $m \in R^\times.$ 47. Let $V$ be a vector space $\ne 0$ over a field $F.$ Then (i) $\GL(V) \le \Aut V^+.$ (See 2.15, 2.16.) (ii) If $F = \mathbb Z_p$ and $V$ has finite dimension $n$ over $\mathbb Z_p$ then $\Aut V^+ \cong \GL_n(\mathbb Z_p).$ ***Sol.*** (i) If $\alpha \in \GL(V),$ then for each $u, v \in V,$ $(u + v)\alpha = u\alpha + v\alpha,$ by linearity, and $\alpha$ is invertible by definition, hence it is an automorphism of $V^+.$ (ii) Suppose $V$ has finite dimension over $F = \mathbb Z_p.$ Let $\alpha \in \Aut V^+.$ Then for any $u, v \in V,$ $(u + v)\alpha = u\alpha + v\alpha$, which implies that for any $a, b \in \mathbb Z_p,$ considering $a$ and $b$ as integers, we have $(au + bv)\alpha = a(u\alpha) + b(v\alpha).$ Hence, $\alpha \in \GL(V).$ From (i), this implies that $\Aut V^+ = \GL(V),$ and we know that $\GL(V) \cong \GL_n(\mathbb Z_p).$ 48. If $G_1 \cong G_2$ then $\Aut G_1 \cong \Aut G_2$ and $\Inn G_1 \cong \Inn G_2.$ ***Sol.*** Let $\varphi \colon G_1 \to G_2$ be an isomorphism. Given an automorphism $\alpha_1$ of $G_1$, define $\alpha_2 \colon G_2 \to G_2$ as $\alpha_2 = \varphi^{-1}\alpha_1\varphi.$ Then $\alpha_2,$ being a composition of isomorphisms, is itself an isomorphism, and hence is an automorphism of $G_2.$ Now, define $f \colon \Aut G_1 \to \Aut G_2$ as $\alpha_1 f = \alpha_2,$ for each $\alpha_1 \in \Aut G_1,$. Then $f$ is an isomorphism: For each $\alpha_1, \beta_1 \in \Aut G_1,$ \begin{align*} (\alpha_1 \beta_1)f & = \varphi^{-1} (\alpha_1 \beta_1) \varphi \\ & = (\varphi^{-1} \alpha_1 \varphi)(\varphi^{-1} \beta_1 \varphi) \\ & = (\alpha_1 f)(\beta_1 f) \end{align*} and the map $g \colon \Aut G_2 \to \Aut G_1$ defined by $\alpha_2 g = \varphi \alpha_2 \varphi^{-1},$ for all $\alpha_2 \in \Aut G_2,$ is an inverse of $f.$ Hence, $\Aut G_1 \cong \Aut G_2.$ Now consider the restriction of $f$ to $\Inn G_1.$ Let $\alpha_1 \in \Inn G_1,$ and $\alpha_2 = \alpha_1 f.$ Thus, there exists $g \in G,$ such that for each $x \in G_1,$ $x\alpha_1 = g^{-1} x g$. Let $h = g\varphi,$ and observe that $(x\alpha_1)\varphi = (g\varphi)^{-1} (x\varphi) (g\varphi) = h^{-1} (x\varphi) h.$ From this it follows that for any $y \in G_2,$ $y\alpha_2 = h^{-1} y h.$ Thus, $\alpha_2 \in \Inn G_2.$ Also, if $\alpha_2$ is any inner automorphism of $G_2,$ with say, $y\alpha_2 = h^{-1} y h,$ for some $h \in G_2,$ for all $y \in G_2,$ then clearly, $\alpha_2 = \alpha_1 f,$ where $\alpha_1$ is defined by $x\alpha_1 = g^{-1} x g,$ for all $x \in G_1,$ with $g = h\varphi^{-1}.$ Thus, the restriction of $f$ to $\Inn G_1$ is surjective onto $\Inn G_2,$ and is injective as well, since $f$ is injective. Hence, this map is an isomorphism from $\Inn G_1$ to $\Inn G_2.$ 49. Define a relation $\sim$ on $G$ by \begin{equation*} x \sim y \quad \text{if and only if} \quad g^{-1} x g = y \quad \text{for some}\ g \in G. \end{equation*} Show that this relation $\sim$ of *conjugacy* is an equivalence relation on $G.$ ***Sol.*** Let $x, y, z \in G.$ Since $1^{-1} x 1 = x,$ $x \sim x.$ If $x \sim y,$ then $g^{-1} x g = y,$ for some $g \in G.$ This implies that $(g^{-1})^{-1} y (g^{-1}) = x,$ and hence $y \sim x.$ If $x \sim y$ and $y \sim z$, then $g^{-1} x g = y$ and $h^{-1} y h = z,$ for some $g, h \in G.$ Conjugating the former equation by $h,$ $h^{-1} g^{-1} x g h = h^{-1} y h$ $\implies$ $(gh)^{-1} x (gh) = z,$ and hence $x \sim z.$ Thus, $\sim$ is reflexive, symmetric, and transitive. 50. If $\alpha \in \Aut G$ and $x \in G$ then \begin{equation*}|x^\alpha| = |x|. \end{equation*} In particular, conjugate elements of a group have the same order. ***Sol.*** If $x^n = 1$, then $(x^\alpha)^n = (x^n)^\alpha = 1^\alpha = 1.$ This implies that $|x^\alpha|$ divides $|x|.$ But $\alpha \in \Aut G$ $\implies$ $\alpha^{-1} \in \Aut G$ and $x = (x^\alpha)^{\alpha^{-1}},$ hence $|x|$ divides $|x^\alpha|$ as well. Thus, $|x| = |x^{\alpha}|.$ 51. Find a group $G$ with a subgroup $K$ and element $g$ such that $g^{-1}Kg \ne K.$ ***Sol.*** Let $G = \Sigma_3,$ $K = \{(), (12)\},$ and $g = (13).$ Now, $g^{-1}(12)g = (13)(12)(13) = (23).$ Hence, $g^{-1}Kg = \{(), (23)\} \ne K.$ 52. (i) The map defined by $g \mapsto g^{-1}$ for all $g \in G$ is an automorphism of $G$ if and only if $G$ is Abelian. (ii) If $G$ is any group for which $g^2 \ne 1$ for some $g \in G$ then $\Aut G \ne 1.$ (Remark. In fact $\Aut G \ne 1$ if and only if $|G| > 2.$ The proof is completed by observing that if $g^2 = 1$ for all $g \in G,$ so that by <a href="#Q3">3</a> $G$ is Abelian, then $G$ has the structure of a vector space $V$ over $\mathbb Z_2,$ and any invertible linear map $V \to V$ is an automorphism of $G.$ For the finite case, see <a href="#Q42">42</a>.) ***Sol.*** (i) The map is self-inverse, and is therefore a bijection. It is a homomorphism if and only if, for all $x, y \in G,$ $(xy)^{-1} = x^{-1} y^{-1}.$ Equivalently (taking inverse on both sides), $xy = (x^{-1} y^{-1})^{-1} = yx,$ that is, $G$ is Abelian. (ii) If $G$ is Abelian, then the map $x \mapsto x^{-1}$ is an automorphism, and it is non-trivial, since $g^2 \ne 1$ $\implies$ $g \ne g^{-1}.$ If $G$ is non-Abelian, then let $x$ and $y$ be elements of $G$ that do not commute. Then the inner automorphism $z \mapsto x^{-1} z x$ is a non-trivial automorphism, since $y \ne x^{-1} y x.$ Thus, in either case, $\Aut G \ne 1.$ 53. (i) Let $\alpha \in \Aut G$ and let \begin{equation*} H = \{\, g \in G \mid g^\alpha = g \,\}. \end{equation*} Prove that $H$ is a subgroup of $G:$ it is called the *fixed point subgroup of $G$ under $\alpha.$* (ii) Let $n$ be a positive integer and $F$ a field. For any $n \times n$ matrix $y$ with entries in $F,$ let $y'$ denote the transpose of $y.$ Show that the map \begin{equation*} \ * \colon \GL_n(F) \to \GL_n(F), \end{equation*} defined by \begin{equation*} \ * \colon x \mapsto (x^{-1})' \end{equation*} for all $x \in \GL_n(F),$ is an automorphism of $\GL_n(F),$ and that the corresponding fixed point subgroup consists of all orthogonal $n \times n$ matrices with entries in $F$ (that is, matrices $y$ such that $y'y = 1$). Prove (by assuming the contrary and considering determinants) that $*$ is an outer automorphism of $\GL_n(F)$ if $F \ne \mathbb Z_2$ and $F \ne \mathbb Z_3.$ (Remark. In fact, $*$ is an outer automorphism of $\GL_n(F)$ unless either $F = \mathbb Z_2$ and $n \le 2$ or $F = \mathbb Z_3$ and $n = 1$). ***Sol.*** (i) $H \ne \varnothing,$ since $1^\alpha = 1$ $\implies$ $1 \in H.$ If $x, y \in H,$ then $x^\alpha = x$ and $y ^\alpha = y$ $\implies$ $(xy^{-1})^\alpha = x^\alpha (y^\alpha)^{-1} = xy^{-1}$ $\implies$ $xy^{-1} \in H.$ Thus, $H \le G.$ (ii) Since both transposition and inversion preserve invertibility and are self-inverse maps, their composition, $*$, is a bijection from $\GL_n(F)$ to itself. To see that it is a homomorphism, observe that for all $x, y \in \GL_n(F),$ \begin{equation*} (xy)^* = ((xy)^{-1})' = (y^{-1} x^{-1})' = (x^{-1})' (y^{-1})' = x^* y^*. \end{equation*} Thus, $* \in \Aut \GL_n(F).$ Now, let $x \in \GL_n(F).$ Then $x$ is a fixed point of $*$ if and only if $x = x^* = (x^{-1})' = (x')^{-1},$ which is equivalent to $x'x = 1.$ Thus, $x$ is a fixed point of $*$ if and only if it is an orthogonal matrix over $F$ (note that every orthogonal matrix over $F$ is an element of $\GL_n(F)$). Now we prove that $*$ cannot be an inner automorphism. Suppose, to the contrary, that $*$ is an inner automorphism. That is, suppose that there exists $g \in GL_n(F)$ such that for all $x \in GL_n(F),$ $x^* = g^{-1} x g.$ Then \begin{equation*} \det x^* = \det(g^{-1} x g) = (\det g)^{-1}(\det x)(\det g) = \det x. \end{equation*} But $\det x^* = \det ((x^{-1})') = (\det x)^{-1}.$ If $F \ne \mathbb Z_2$ and $F \ne \mathbb Z_3,$ then $F$ has a non-zero element $k$ such that $a \ne a^{-1},$ and there exists $x \in \GL_n(F)$ with $\det x = a$ (for example, the diagonal matrix with $x_{11} = a$ and $x_{ii} = 1$ for all $i > 1$). Thus, for this matrix $x,$ $\det x \ne (\det x)^{-1},$ which is a contradiction. Therefore, $*$ is not an inner automorphism. 54. Let $\alpha \in \Aut G.$ Then $\alpha$ is said to be *fixed-point-free* if the fixed point subgroup of $G$ under $\alpha$ is trivial (see <a href="#Q53">53</a>); that is, if $g^\alpha \ne g$ whenever $1 \ne g \in G.$ (Remark. The term 'fixed-point-free' is standard. It is perhaps a slight abuse of language, since of course any automorphism of a group fixes the identity element. To say that an automorphism is fixed-point-free means that it fixes no element of the group other than the identity element.) (i) Suppose that $\alpha$ is a fixed-point free automorphism of the finite group $G.$ Show that \begin{equation*} \{\, g^\alpha g^{-1} \mid g \in G \,\} = G. \end{equation*} Deduce that if $|\alpha| = 2$ then, for all $x \in G,$ \begin{equation*} x^{\alpha} = x^{-1}, \end{equation*} and that $G$ is Abelian of odd order greater than $1$. (ii) Let $G$ be a finite Abelian group of odd order greater than $1.$ Then the map \begin{equation*} \alpha : x \to x^{-1}, \end{equation*} defined for all $x \in G,$ is a fixed-point-free automorphism of $G$ of order $2$. (Hints. For (i), show that the map of $G$ to itself defined, for all $g \in G,$ by $g \mapsto g^\alpha g^{-1},$ is injective. Then use <a href="#Q52">52 (i)</a> and 1.13.) ***Sol.*** (i) Let $\varphi \colon G \to G$ be the map $g\varphi = g^{\alpha}g^{-1},$ for all $g \in G.$ Then $\varphi$ is injective: if $g\varphi = h\varphi,$ then $g^{\alpha}g^{-1} = h^{\alpha}h^{-1}$ $\implies$ $(h^{-1}g)^\alpha = h^{-1}g$ $\implies$ $h^{-1}g = 1,$ since $\alpha$ is fixed-point-free, and therefore $g = h.$ Since $G$ is finite, $\varphi$ is bijective, and therefore $G = \Im \varphi = \{\, g^\alpha g^{-1} \mid g \in G \,\}.$ Now, let $x \in G.$ Then $x = g^{\alpha}g^{-1},$ for some $g \in G,$ which implies that \begin{equation*} x^\alpha = g^{\alpha^2} \left( g^{-1} \right)^\alpha = g (g^\alpha)^{-1} = \left( g^\alpha g^{-1} \right)^{-1} = x^{-1}. \end{equation*} Now, from <a href="#Q52">52</a>, $G$ must be Abelian, since $\alpha$ is an automorphism of $G.$ Also, $|G| > 1,$ for otherwise $\alpha$ would be trivial, whereas we know that $|\alpha| = 2.$ To see that $|G|$ is odd, observe that all the elements of $G$ that are not self-inverse can be paired up with their respective inverses, and these together constitute an even number of elements of $G.$ Since $\alpha$ is fixed-point-free, the only self-inverse element of $G$ is $1,$ and therefore the total number of elements of $G$ is odd. (ii) From <a href="#Q52">52</a>, we know that if $G$ is Abelian, then $\alpha \in \Aut G.$ To see that $\alpha$ is fixed-point-free, let $x$ be a fixed point of $\alpha.$ Then $x^\alpha = x^{-1} = x$ $\implies$ $x^2 = 1.$ Thus, either $x = 1$ or $|x| = 2.$ But since $|G|$ is odd, $|x|$ cannot be even, hence the only fixed point of $\alpha$ is $1.$ 55. Let $X$ be any non-empty set. Let $c$ be any positive real number and define, for all $x, y \in X,$ \begin{equation*} d(x,y) = \begin{cases} 0 & \text{if $x = y$} \\ c & \text{if $x \ne y$.} \end{cases} \end{equation*} Show that $d$ is a distance function for $X,$ and that for this metric space \begin{equation*} \Isom X = \Sigma_X. \end{equation*} ***Sol.*** By definition, $d(x, y) = 0$ if and only if $x = y,$ and $d(x, y) = d(y, x),$ for all $x, y \in X.$ Now, let $x, y, z \in X.$ If $x = z,$ then $0 = d(x, z) \le d(x, y) + d(y, z).$ If $x \ne z,$ then $y$ is distinct from at least one of $x$ and $z,$ and hence $c = d(x, z) \le d(x, y) + d(y, z).$ Therefore, $d$ is a distance function on $X$. Now, let $\sigma \in \Sigma_X.$ Then for any $x, y \in X,$ $x \ne y \iff x\sigma \ne y\sigma,$ which implies that $d(x, y) = c \iff d(x\sigma, y\sigma) = c,$ and $d(x, y) = 0 \iff d(x\sigma, y\sigma) = 0.$ Hence, $d(x\sigma, y\sigma) = d(x, y).$ Therefore, $\sigma \in \Isom X.$ 56. View $\mathbb R$ as the Euclidean line $E^1$ with the usual distance function (that is, for $x, y \in \mathbb R,$ $d(x, y) = |x - y|$). (i) For each $a \in \mathbb R,$ the map \begin{equation*} \tau_a \colon x \mapsto x + a \qquad \text{for all $x \in \mathbb R$} \end{equation*} is an isometry of $\mathbb R,$ called a *translation*, and \begin{equation*} \mathbb R^+ \cong T = \{\, \tau_a \mid a \in \mathbb R \,\} \le \Isom \mathbb R. \end{equation*} (ii) For each $a \in \mathbb R,$ the map \begin{equation*} \varepsilon_a \colon x \to a - x \qquad \text{for all $x \in \mathbb R$} \end{equation*} is an isometry of $\mathbb R$ called a *reflexion*, and \begin{equation*} \varepsilon^2_a = 1\,(= \tau_0) \qquad \text{and} \qquad \varepsilon_a \tau_b = \tau_{-b} \varepsilon_a \qquad \text{for all $a, b \in \mathbb R$.}\end{equation*} (iii) Every isometry of $\mathbb R$ is either a translation or a reflexion. (iv) $\Isom \mathbb R$ is a non-Abelian group and $T$ is an Abelian subgroup of index $2.$ ***Sol.*** (i) For all $a \in \mathbb R$, $d(x \tau_a, y \tau_a)$ $=$ $d(x + a, y + a)$ $=$ $|(x + a) - (y + a)|$ $=$ $|x - y|$ $=$ $d(x, y)$. Thus, $\tau_a$ is an isometry of $\mathbb R$. Define $\tau \colon R \to \Isom \mathbb R$ by $a \mapsto \tau_a$. If $a, b \in \mathbb R$, then $\tau(a + b) = \tau_{a + b}$ is the isometry that maps each $x \in \mathbb R$ to $(a + b) + x$ $=$ $(x + a) + b$ $=$ $(x \tau_a)\tau_b$. Hence, $\tau(a + b)$ $=$ $\tau_a \tau_b$ $=$ $\tau(a) \tau(b)$, and therefore $\tau$ is a homomorphism from $\mathbb R^+$ to $\Isom \mathbb R$. Now, if $a, b \in \mathbb R$ are such that $\tau(a) = \tau(b)$, then for $a$ = $0 + a$ $=$ $0\tau_a$ $=$ $0\tau(a)$ $=$ $0\tau_b(0)$ $=$ $0\tau_b$ $=$ $0 + b = b$. Thus, $\tau$ is injective. Clearly, $\Im \tau = \{\, \tau_a \mid a \in \mathbb R \,\}$ $\le$ $\Isom \mathbb R$. (ii) For all $a \in \mathbb R$, $d(x \varepsilon_a, y \varepsilon_a)$ $=$ $d(a - x, a - y)$ $=$ $|(a - x) - (a - y)|$ $=$ $|y - x|$ $=$ $d(x, y)$. Thus, $\varepsilon_a$ is an isometry of $\mathbb R$. Since, $(x \varepsilon_a)\varepsilon_a$ $=$ $(a - (a - x))$ $=$ $x$, $\varepsilon_a^2 = 1$. Also, $(x \varepsilon_a)\tau_b$ $=$ $(a - x)\tau_b$ $=$ $a - x + b$ $=$ $a - (x - b)$ $=$ $(x + (-b))\varepsilon_a$ $=$ $(x \tau_{-b})\varepsilon_a$. Hence, $\varepsilon_a \tau_b = \tau_{-b} \varepsilon_a$. (iii) We know that for all $a \in \mathbb R$, $\tau_a, \varepsilon_a \in \Isom \mathbb R$. We will show that every isometry of $\mathbb R$ is either $\tau_a$ or $\varepsilon_a$ for some $a \in R$. Let $\varphi$ be any isometry of $\mathbb R$, and let $a = 0\varphi$. Now, if $x \in \mathbb R$, $d(x, 0) = d(x\varphi, 0\varphi)$ implies that $|x| = |x\varphi - a|$, and hence $x\varphi = a + x$ or $x\varphi = a - x$. That is, $\varphi = \tau_a$ or $\varphi = \varepsilon_a$. Therefore, every isometry of $\mathbb R$ is either a translation or a reflexion. (iv) We know from (ii) that, for example, $\varepsilon_1 \tau_1 = \tau_{-1} \varepsilon_1$, and $\tau_{-1} \ne \tau_1$. Hence, $\Isom \mathbb R$ is non-Abelian. From (i), $T$ is a group isomorphic to $\mathbb R^+$, and hence is an Abelian subgroup of $\Isom \mathbb R$. To show that the index of $T$ in $\Isom \mathbb R$ is $2$, we will show $T$ and $\varepsilon_0 T$ are the only two left cosets of $T$ in $\Isom \mathbb R$. We know that every isometry of $\mathbb R$ is etiher $\tau_a \in T$ or $\varepsilon_a$, for some $a \in \mathbb R$. Hence, every element of $\Isom \mathbb R$ not in $T$ is of the form $\varepsilon_a$, for some $a \in \mathbb R$. But $\varepsilon_a = \varepsilon_0 \tau_a \in \varepsilon_0 T$. 57. Let the notation be as in <a href="#Q56">56</a>. Show that for every $n \in \mathbb Z,$ \begin{equation*} \tau_1^n = \tau_n, \end{equation*} and that the elements of the symmetry group $S_{\mathbb R}(\mathbb Z)$ are just the isometries \begin{equation*} \tau_1^n \qquad \text{for all $n \in \mathbb Z$} \qquad \text{and} \qquad \tau_1^n \varepsilon_0 \qquad \text{for all $n \in \mathbb Z$.} \end{equation*} Note that $\varepsilon_0^2 = 1$ and $\varepsilon_0 \tau_1 = \tau_1^{-1} \varepsilon_0.$ The group $S_{\mathbb R}(\mathbb Z)$ is called *the infinite dihedral group* and denoted by $D_\infty.$ 58. Let $n$ be an integer, $n \ge 3.$ Then $D_{2n}$ can be embedded in $\Sigma_n.$ Moreover, $D_6 \cong \Sigma_3$ but, whenever $n > 3$, $D_{2n} \not\cong \Sigma_n.$ (It may be assumed that an isometry of $E^2$ is uniquely determined by its effect on any 3 non-collinear points.) 59. Let $n$ be an integer, $n \ge 3,$ and let $L = D_{2n}.$ Let $J = \langle \rho \rangle,$ where $\rho$ is defined as in 2.24. Show that every element of $L \setminus J$ has order $2.$ Deduce that $J$ is the only cyclic subgroup of $L$ of order $n.$ 60. Every group of order $6$ is isomorphic to either $C_6$ or $D_6.$ Hence, in the notation of chapter 1, $\nu(6) = 2.$ (Hints. Let $G$ be a non-cyclic group of order $6.$ By <a href="#Q42">42</a>, $G$ has an element $x$ of order $3.$ Let $y$ be an element of $G$ other than $1, x, x^2.$ Then $G = \{1, x, x^2, y, xy, x^2 y\}$, $y^2 = 1$ and $yx = x^2 y.$) 61. (i) For any 2 points $s,$ $t$ of $E^2$ there is a unique translation $\tau_a$ of $E^2$ which maps $s$ to $t.$ (ii) For any 2 points $s,$ $t$ of $E^2,$ if $\tau_a$ is the unique translation of $E^2$ which maps $s$ to $t$ then \begin{equation*} \tau_a^{-1} \Rot(E^2; s) \tau_a = \Rot(E^2; t). \end{equation*} Thus any 2 groups of rotations are conjugate subgroups of $\Isom E^2.$