Group Theory
, Groups
, Group Actions
, Solutions
Any group of prime order is cyclic.
Sol. The only divisors of a prime number are and the number itself, and therefore the order of any subgroup must be equal to 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 is the least prime number) generates a non-trivial cyclic subgroup, which is necessarily equal to the whole group.
Any two cyclic groups of the same finite order are isomorphic.
Sol. Let and be two cyclic groups of finite order Define by for all Then is well-defined and injective, since if and only if which is equivalent to For any two elements Hence, is an injective homomorphism from to and therefore is an isomorphism, since
If for every then is an Abelian group.
Sol. For all Now, for any
Let Then
Sol. First, observe that for any Now, for any Hence,
(i) Let with Then for every integer
(ii) If is a cyclic group of finite order then the number of distinct elements which generate is where is Euler's totient function: that is, is the number of positive integers not exceeding which are co-prime to
Sol. (i) Let Since is an integer, and it follows that and hence On the other hand, suppose that for some integer Then Thus, is the least positive integer such that
(ii) Let where For any element
Let with If and are co-prime and and commute, then
Sol. First, On the other hand, suppose is a positive integer such that Then Hence, which implies that Since and are co-prime, this is possible only if and which implies that and Therefore, . Thus, is the least positive integer such that
Let with where and are co-prime positive integers. Then there are elements such that and Moreover, and are uniquely determined by these conditions.
Sol. Since and are co-prime, there exist integers and such that Let and Then,
since any two powers of commute. Also, implies that and are co-prime. Therefore, Similarly,
Now, suppose satisfy and Then Similarly, Hence, and are uniquely determined by the given conditions.
By considering orders of elements in the multiplicative group of all non-zero elements of the field prove Fermat's theorem: for every integer not divisible by
Sol. The multiplicative group of non-zero elements of has order If is not divisible by then and are co-prime, and hence where with and co-prime. Hence, is in the mutliplicative group, which implies that Therefore,
Let be an Abelian group of finite order Show that the product of the distinct elements of is equal to the product of all elements of of order (where the latter product is interpreted as if has no element of order ). By applying this result to the multiplicative group of all non-zero elements of the field prove Wilson's theorem for prime numbers:
Sol. Let For each with if and only if Therefore, in the product every non-identity element of order other than cancels out with its inverse. Hence, this product is equal to the product of the elements of order
Now, let be the multiplicative group of the non-zero elements of namely Then is the product of all elements of If has order then there exists a positive integer such that But and are co-prime to and therefore must be equal to Hence, is the only element of having order Thus,
Let and let Then if and only if
Sol. Recall that is equivalent to and that is equivalent to Now, that is, which is equivalent to
(i) Let Then is finite if and only if and are both finite, and if so,
(ii) Let with prime. Then either or
Sol. (i) Let be the distinct representatives of left cosets of in and let be the disinct representatives of in
Observe that if then since Also, if and are distinct, then they are disjoint, which implies that for any Thus, are distinct representatives of left cosets of in It remains to show that these are form an exhaustive system of distinct representatives of left cosets of in But this is obvious, since, for any for some and and then for some so
Now, since (as sets), it follows that 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 either or In the former case, and in the latter,
Let and If and where and are co-prime integers, then
Sol. There exist integers and such that Hence, (since ).
Give an example of a semigroup without an identity element.
Sol. The set of natural numbers, under addition.
Give an example of an infinite semigroup with an identity element such that no element except has an inverse.
Sol. The set under multiplication.
Let be a semigroup and let Show that forms a subgroup of (of order ) if and only if Such an element is called an idempotent in (Warning. A semigroup may have several different subgroups of order : see 16. Why does a group have only one subgroup of order ?)
Sol. If is a subgroup of then Conversely, if then is an identity element in Since this is unique element of is a group, and hence is a subgroup of
A group has only one subgroup of order which is since for any element of a group, by cancellation.
Let be any non-empty set. Let be the set of all subsets of Show that is a semigroup with respect to the operation Does have an identity element, and if so, what are the units in ? Show that every element of is an idempotent (15). Deduce that for all is a subgroup of and that every subgroup of has order
What happens if is replaced by ?
Sol. The intersection of any two subsets of is a subset of and is associative, hence is an associative binary operation on which shows that is a semigroup.
For any Hence, (which is a subset of itself, and is therefore in ) is the identity element of
If then for all which shows that cannot be a unit. Thus, is the only unit of
The intersection of any set with itself is itself, hence every element of is idempotent. Therefore, for any implies that is a subgroup of
A group can only have one idempotent elements, but since all elements of are idempotent, no two of them can be in the same subgroup of Thus, all subgroups of have order
If is replaced by all the above results hold, except that the identity element becomes instead of
Let be a set with Do two elements of necessarily commute? Do two elements of necessarily commute?
Note. is the semigroup of all endofunctions of and is its group of units.
Sol. No, not all elements of mutually commute. Let let be the endofunction that maps both elements of and let be the endofunction that maps both elements to Then
and therefore any two elements of commute (since, e.g., every element commutes with itself and with the identity).
Express the permutation as a product of disjoint cycles. Find expressions for and as products of disjoint cycles.
Sol. and
If and are homomorphisms, then the composite map is also a homomorphism.
Sol. For all
Verify that if is any set of groups then is an equivalence relation on
Sol. The identity map of any group is an isomorphism from to itself, hence If then there is an isomorphism and an inverse isomorphism which implies that Finally, if and then there exist isomorphisms and Then is a homomorphism from to and since the composition of bijections is a bijection, is an isomorphism. Hence,
(i) Calculate the products and in the group where is an integer with
(ii) A permutation such as which interchanges two points and fixes all others, is called a transposition. For any integer show that the permutation can be expressed as a product of transpositions.
(iii) For any integer prove that every non-trivial element of can be expressed as a product of at most transpositions.
Sol. (i) and
(ii) by induction on
(iii) We know that any non-trivial element of can be expressed as a sum of disjoint cycles, such that the sum of the lengths of the cycles (including cycles of length ) is From (ii), any cycle of length can be expressed as a product of transpositions. Thus, any non-trivial element of can be expressed as product of transpositions, where is the number of cycles in its cycle decomposition.
Let be a positive integer and let Suppose that can be expressed as a product of disjoint cycles of lengths respectively, where are positive integers such that Then is the least common multiple of
Sol. If a permutation is a cycle of length then its order is Let be the cycles in the cycle decomposition of Since disjoint cycles commute, Moreover, if and are disjoint, then so are and and hence, for each Thus, is a multiple of for all Then being the least such positive integer, is the least common multiple of
Let and be positive integers such that divides Let be a cycle of length in Then is the product of disjoint cycles of length
Sol. Let and without loss of generality, let Now, Hence, (say) is a cycle of Next, so is a cycle of Proceeding similarly, is a cycle of and Thus the cycle decomposition of is with each cycle having length
Any group which can be embedded in an Abelian group is Abelian.
Sol. Let be an embedding of into an Abelian group Then for any and since is injective, this implies that Therefore, is Abelian.
Prove that if is a non-empty set and a non-empty subset of then can be embedded in (In particular, whenever and are positive integers such that then can be embedded in ) Hence or otherwise show that if then is non-Abelian.
Sol. Define a homomorphism as follows: For each let be the permutation of defined by
To see that is a homomorphism, first note that Now, let and let
If then
If then
Thus,
Moreover, is injective: for any and if then Therefore, is an embedding of in
Now, if then let have three elements. Then is non-Abelian, and since embeds in the latter is non-Abelian too (see 24).
If is a homomorphism and is Abelian then is Abelian.
Sol. Let Then
Suppose that and are isomorphic groups, and let be an isomorphism. If and then
Sol. Let be a system of distinct representatives of left cosets of in For each left coset is a left coset of in If then and therefore (since is injective), which implies that Thus, maps distinct left cosets of to distinct left cosets of Since is surjective, any element of any coset of can be written as for some and then for some Thus, which shows that every left coset of in is of the form for some Hence, the image of under is a system of distinct representatives of left cosets of in and is in bijection with Therefore,
If and are finite, then a necessary condition that can be embedded in is that divides Show by example that this is not a sufficient condition.
Sol. We know that if can be embedded in then is isomorphic to a subgroup of Thus, must be the order of a subgroup of which implies that divides To see that this condition is not sufficient, let which is of order and let be the cyclic group of order Then but which is non-Abelian, cannot be embedded in which is Abelian (see 24).
(i) Any infinite cyclic group is isomorphic to
(ii) If is a non-trivial group of which the only subgroups are and then is cyclic of prime order.
Sol. Let be an infinite cyclic group. Define by for each Note that if and only if which is equivalent to since has infinite order. Thus, is well-defined and injective. Finally, for any hence is surjective. Thus, is an isomorphism from to
(ii) Let be non-trivial. Then being a non-trivial subgroup of must be equal to Now, let by any factorisation of into two positive integers. Then is a subgroup of having order which implies that or That is, the only divisors of are and itself, which shows that () is prime.
(i) If is finite then no proper subgroup of is isomorphic to
(ii) If is a group such that whenever then every element of has finite order. (An example of an infinite group with this property will be given in 144.)
(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 having infinite order. Let and Then but which contradicts the given condition.
Show that has infinitely many distinct subgroups. Deduce that every infinite group has infinitely many distinct subgroups.
Sol. For each we have a subgroup For any integer if and only if for some integer i.e., Thus, only if and which implies that (since ). Thus, is a collection of infinitely many distinct subgroups of
Now, let be any infinite group. If contains an element, say of infinite order, then contains infinitely many distinct subgroups, all of which are subgroups of
On the other hand, suppose every element of has finite order. Then consider the collection of cyclic subgroups generated by all the elements of If then, since is finite, there are infinitely many such that Thus, has infinitely many distinct subgroups.
No two of the groups are isomorphic. (Remark. It is in fact true that This can be proved by regarding and as vector spaces over and using vector space theory and facts about infinite cardinals.)
Sol. Let be any injective homomorphism, and let Then, but there exists no such that since otherwise would imply that (by injectivity of ), but contains no such Therefore, is not surjective. This shows that there is no isomorphism from to Since it follows that cannot be isomorphic to either.
Next, suppose that is an isomorphism. For any implies that It follows that if then Now, let be such that and let be such that Then 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 is countable and is uncountable, and hence there is no bijection between them, let alone an isomorphism of groups.
Let denote the set of all homomorphisms of a group into an Abelian group Define a binary operation on as follows: for
is defined by
for all Verify that and that, with respect to acquires the structure of an Abelian group.
Show that for any Abelian group
Sol. Let Then, for any two elements
Thus,
Associativity and commutativity of in follow from associativity and commutativity of multiplication in The trivial homomorphism from to is the clearly identity element of For any the map defined by for all is a homomorphism from to and is an inverse with respect to as shown below:
which implies that is the trivial homomorphism from to the identity element of
Therefore, is an Abelian group with respect to
Now, for each let be the function defined by for all Then is a homomorphism from to since
Next, define a map by for all
First, observe that is a homomorphism: if and To see that these two maps are equal, let Then
Next, to show that is bijective, observe that every element of is uniquely determined by the value of : if then which shows that every is of the form with and that Therefore, is an isomorphism from to
Show that if is a homomorphism, and is a positive integer such that then Hence, by considering the elements of satisfying or otherwise, show that is not isomorphic to either or
Sol.
Now, let be any homomorphism, where or Let Then But the only such element in is hence which shows that is not injective, and hence cannot be an isomorphism.
Prove that is not isomorphic to (Hint. Note that for any positive real number and any positive integer there is a unique positive real number such that )
Sol. Suppose, to the contrary, that there exists an isomorphism Then, since and are the only square roots of in and is injective. This implies that for all
Now, let If then which is impossible, as If then which is impossible, as Thus, there is no such isomorphism
Prove that there is no field such that (Hint. Assume that there is a field with an isomorphism and consider )
Sol. If possible, let be an isomorphism, and let Then, This implies that (for, if then ), and therefore, Now, for all But this implies Hence, and therefore cannot be isomorphic to
Let be a field. Define a binary operation on by
Prove that the set of all elements of distinct from forms a group with respect to the operation and that
Sol. Let
Observe that if then Therefore, if then and are in and From this, it is easy to see that is an assocative and commutative binary operation on and that is the identity element and for each if i.e., (which is clearly well-defined and not equal to ), so every element of has an inverse with respect to Therefore, is a group.
Now, define by for all This is well defined, since It is also obvious that is bijective, with inverse for all The earlier observation shows that hence is an isomorphism.
The set of all positive rational numbers forms a subgroup of and there is a surjective homomorphism
which is not an isomorphism.
Is ?
Sol. For any two rational numbers and which shows that is a homomorphism. For each positive rational number hence is a surjection from to Finally, and hence is not an isomorphism.
To see that suppose, to the contrary, that there exists an isomorphism Let Then for some But which is impossible, since Hence,
For any positive integer the only homomorphism is the trivial homomorphism.
Sol. Let be any homomorphism, and let Then, Since every element of is a power of this implies that maps all elements of to and is therefore the trivial homomorphism.
Let be a positive integer. Then
(i)
(ii) is an Abelian group of order where is Euler's function (see 5).
(iii) By considering orders of elements in prove the Euler-Fermat theorem:
whenever is an integer co-prime to (This generalizes 8).
Sol. (i) Let and define by for all Then for since whenever Thus, Since and is clearly surjective, is an isomorphism.
(ii) Multiplication is associative and commutative in and all elements of are invertible, by definition. If then hence Therefore, is an Abelian group.
Now, let
First, suppose that is not co-prime to so Then is a non-zero element of and (since ). Thus, is a zero-divisor and hence is not a unit of
Next, suppose that is co-prime to Then there exist integers and such that Thus, (modulo ) is a multiplicative inverse of and is a unit of Therefore, consists of all the positive integers less than and co-prime to it. Hence,
(iii) From (ii), if is co-prime to then for some Then since
(i) Let be a group and a positive integer such that for all Show that if is a homomorphism then Hence show that (cf. 33).
(ii) Deduce that if is a finite group then so is (Remark. If is a finite Abelian group then in fact but we do not yet have the means of proving this. See 454.)
Sol. (i) For each
Now, let be the inclusion map, which is always an injective homomorphism. As proved above, is surjective as well. Hence,
(ii) If is finite, of order then for all Hence, by (i), which is finite since and are both finite.
Let be a finite group such that for all Show that for some finite dimensional vector space over Deduce that for some integer (Hint. By 3, is Abelian. Let consist of the same elements as with the sum of 2 elements of equal to their product in and scalar multiplication defined in the obvious way.)
Sol. Let Define by for all Then is clearly an Abelian group. Define by for all and
Now, for and
and
Thus, is a vector space over
Let be the function defined by for all From the definition of it is evident that for all Since is obviously a bijection. Hence. is an isomorphim from to Now, is a finite dimensional vector space over (since it is finite), and therefore is isomorphic to for some integer Hence,
Let be a field and let be positive integers with Then
(i) can be embedded in
(ii) is non-Abelian for all
Sol. (i) Let Define a map by
where and are zero matrices of orders and respectively, and is the identity matrix of order
From the laws of multiplication of partitioned matrices, it is clear that is a homomorphism. Since equality of two matrices implies equality of all corresponding pairs of submatrices, is an embedding.
(ii) Let and Then and which implies that is non-Abelian. From (i), can be embedded in for all and therefore, is non-Abelian as well (see 24).
Sol. Consider the vector space with the standard basis Then with the matrices acting on the vectors of by left-multiplication. Let and note that any two distinct vectors of are linearly independent. Thus, any mapping of and to a pair of distinct vectors of defines a unique automorphism of and conversely each automorphism defines such a mapping. But such a mapping is equivalent to a permutation of Therefore, every automorphism of corresponds to a unique permutation of the set which implies that This further implies that
Let be the set of all matrices of the form where are real numbers such that Prove that forms a subgroup of and that the set of all elements of in which forms a subgroup of isomorphic to
Find all elements in of order Hence show that the product of two elements of order in can be an element of infinite order.
Sol. The determinant of any matrix in is hence and is clearly non-empty. Now, let and be any two elements of Then
hence Therefore,
Let be the set of all the matrices in whose diagonal elements are both From if then Define as the function that maps each to the matrix in whose -entry is This is clearly an injection, and the preceding observation shows that it is a homomorphism. It is also evident that Hence, and
To find all elements of of order let Then Therefore, if and only if and either or
Let and Then and are of order but is a non-identity element of and hence has infinite order.
(i) If is a ring with a multiplicative identity then can be embedded in (Hint. See 2.11.)
(ii) and for every positive integer
Sol. (i) For each define a function by for all Distributivity implies that is an endomorphism of Since is a unit, it has a multiplicative inverse and for all which implies that is the identity automorphism. Since and this also implies that is the identity automorphism. Thus, is an automorphism of
Now, define a map by for all This is well-defined, as shown above. If then for all That is, and hence is a group homomorphism. If then Hence, is an embedding.
(ii) Let be the ring or Let be the embedding of into defined in (i). Let be any automorphism of and let Note that Since every element of can be written in the form , and hence But This shows that is surjective, and hence bijective. Therefore, for or
Note. To see that observe that, since is an automorphism, which shows that and hence
Let be a vector space over a field Then
(i) (See 2.15, 2.16.)
(ii) If and has finite dimension over then
Sol. (i) If then for each by linearity, and is invertible by definition, hence it is an automorphism of
(ii) Suppose has finite dimension over
Let Then for any , which implies that for any considering and as integers, we have Hence, From (i), this implies that and we know that
If then and
Sol. Let be an isomorphism.
Given an automorphism of , define as Then being a composition of isomorphisms, is itself an isomorphism, and hence is an automorphism of
Now, define as for each . Then is an isomorphism: For each
and the map defined by for all is an inverse of
Hence,
Now consider the restriction of to Let and Thus, there exists such that for each . Let and observe that From this it follows that for any Thus, Also, if is any inner automorphism of with say, for some for all then clearly, where is defined by for all with Thus, the restriction of to is surjective onto and is injective as well, since is injective. Hence, this map is an isomorphism from to
Define a relation on by
Show that this relation of conjugacy is an equivalence relation on
Sol. Let
Since
If then for some This implies that and hence
If and , then and for some Conjugating the former equation by and hence
Thus, is reflexive, symmetric, and transitive.
If and then
In particular, conjugate elements of a group have the same order.
Sol. If , then This implies that divides But and hence divides as well. Thus,
Find a group with a subgroup and element such that
Sol. Let and
Now, Hence,
(i) The map defined by for all is an automorphism of if and only if is Abelian.
(ii) If is any group for which for some then (Remark. In fact if and only if The proof is completed by observing that if for all so that by 3 is Abelian, then has the structure of a vector space over and any invertible linear map is an automorphism of For the finite case, see 42.)
Sol. (i) The map is self-inverse, and is therefore a bijection. It is a homomorphism if and only if, for all Equivalently (taking inverse on both sides), that is, is Abelian.
(ii) If is Abelian, then the map is an automorphism, and it is non-trivial, since If is non-Abelian, then let and be elements of that do not commute. Then the inner automorphism is a non-trivial automorphism, since
Thus, in either case,
(i) Let and let
Prove that is a subgroup of it is called the fixed point subgroup of under
(ii) Let be a positive integer and a field. For any matrix with entries in let denote the transpose of Show that the map
defined by
for all is an automorphism of and that the corresponding fixed point subgroup consists of all orthogonal matrices with entries in (that is, matrices such that ).
Prove (by assuming the contrary and considering determinants) that is an outer automorphism of if and (Remark. In fact, is an outer automorphism of unless either and or and ).
Sol. (i) since If then and Thus,
(ii) Since both transposition and inversion preserve invertibility and are self-inverse maps, their composition, , is a bijection from to itself. To see that it is a homomorphism, observe that for all
Thus,
Now, let Then is a fixed point of if and only if which is equivalent to Thus, is a fixed point of if and only if it is an orthogonal matrix over (note that every orthogonal matrix over is an element of ).
Now we prove that cannot be an inner automorphism. Suppose, to the contrary, that is an inner automorphism. That is, suppose that there exists such that for all Then
But If and then has a non-zero element such that and there exists with (for example, the diagonal matrix with and for all ). Thus, for this matrix which is a contradiction. Therefore, is not an inner automorphism.
Let Then is said to be fixed-point-free if the fixed point subgroup of under is trivial (see 53); that is, if whenever
(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 is a fixed-point free automorphism of the finite group Show that
Deduce that if then, for all
and that is Abelian of odd order greater than .
(ii) Let be a finite Abelian group of odd order greater than Then the map
defined for all is a fixed-point-free automorphism of of order .
(Hints. For (i), show that the map of to itself defined, for all by is injective. Then use 52 (i) and 1.13.)
Sol. (i) Let be the map for all Then is injective: if then since is fixed-point-free, and therefore Since is finite, is bijective, and therefore
Now, let Then for some which implies that
Now, from 52, must be Abelian, since is an automorphism of Also, for otherwise would be trivial, whereas we know that To see that is odd, observe that all the elements of that are not self-inverse can be paired up with their respective inverses, and these together constitute an even number of elements of Since is fixed-point-free, the only self-inverse element of is and therefore the total number of elements of is odd.
(ii) From 52, we know that if is Abelian, then To see that is fixed-point-free, let be a fixed point of Then Thus, either or But since is odd, cannot be even, hence the only fixed point of is
Let be any non-empty set. Let be any positive real number and define, for all
Show that is a distance function for and that for this metric space
Sol. By definition, if and only if and for all Now, let If then If then is distinct from at least one of and and hence Therefore, is a distance function on .
Now, let Then for any which implies that and Hence, Therefore,
View as the Euclidean line with the usual distance function (that is, for ).
(i) For each the map
is an isometry of called a translation, and
(ii) For each the map
is an isometry of called a reflexion, and
(iii) Every isometry of is either a translation or a reflexion.
(iv) is a non-Abelian group and is an Abelian subgroup of index
Sol. (i) For all , . Thus, is an isometry of . Define by . If , then is the isometry that maps each to . Hence, , and therefore is a homomorphism from to . Now, if are such that , then for = . Thus, is injective. Clearly, .
(ii) For all , . Thus, is an isometry of . Since, , . Also, . Hence, .
(iii) We know that for all , . We will show that every isometry of is either or for some . Let be any isometry of , and let . Now, if , implies that , and hence or . That is, or . Therefore, every isometry of is either a translation or a reflexion.
(iv) We know from (ii) that, for example, , and . Hence, is non-Abelian. From (i), is a group isomorphic to , and hence is an Abelian subgroup of . To show that the index of in is , we will show and are the only two left cosets of in . We know that every isometry of is etiher or , for some . Hence, every element of not in is of the form , for some . But .
Let the notation be as in 56. Show that for every
and that the elements of the symmetry group are just the isometries
Note that and The group is called the infinite dihedral group and denoted by
Let be an integer, Then can be embedded in Moreover, but, whenever , (It may be assumed that an isometry of is uniquely determined by its effect on any 3 non-collinear points.)
Let be an integer, and let Let where is defined as in 2.24. Show that every element of has order Deduce that is the only cyclic subgroup of of order
Every group of order is isomorphic to either or Hence, in the notation of chapter 1,
(Hints. Let be a non-cyclic group of order By 42, has an element of order Let be an element of other than Then , and )
(i) For any 2 points of there is a unique translation of which maps to
(ii) For any 2 points of if is the unique translation of which maps to then
Thus any 2 groups of rotations are conjugate subgroups of