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Solutions to Exercises from John S. Rose's A Course on Group Theory

tags: Group Theory, Groups, Group Actions, Solutions

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=g and
    H=h
    be two cyclic groups of finite order
    n.
    Define
    f:GH,
    by
    f(gi)=hi,
    for all
    gG.
    Then
    f
    is well-defined and injective, since
    gi=gj
    if and only if
    ijmodn,
    which is equivalent to
    hi=hj.
    For any two elements
    gi,gjG,
    f(gigj)=f(gi+j)=hi+j=hihj=f(hi)f(hj).
    Hence,
    f
    is an injective homomorphism from
    G
    to
    H,
    and therefore is an isomorphism, since
    |G|=|H|.

  3. If

    g2=1 for every
    gG,
    then
    G
    is an Abelian group.
    Sol. For all
    gG,
    g2=1g=g1.
    Now, for any
    g,hG,
    gh=(gh)1=h1g1=hg.

  4. Let

    g1,g2G. Then
    |g1g2|=|g2g1|.

    Sol. First, observe that for any
    x,yG,
    xy=1yx=1.
    Now, for any
    nN,
    (g1g2)n=g1(g2g1)n1g2.
    Hence,
    (g1g2)n=1
    g1(g2g1)n1g2=1
    (g2g1)n1g2g1=1
    (g2g1)n=1.

  5. (i) Let

    gG with
    |g|=n<.
    Then for every integer
    m,
    |gm|=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
    φ(n),
    where
    φ
    is Euler's totient function: that is,
    φ(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
    gn=1,
    it follows that
    mkn,
    and hence
    (gm)k=1.
    On the other hand, suppose that
    (gm)l=1,
    for some integer
    l.
    Then
    |g|=nnmlkml/gcd(m,n)kl.
    Thus,
    k
    is the least positive integer such that
    (gm)k=1.

    (ii) Let
    G=g,
    where
    |g|=n.
    For any element
    gmG,
    G=gm
    |gm|=n
    n/gcd(m,n)=n
    gcd(m,n)=1.

  6. Let

    g1,g2G with
    |g1|=n1<,
    |g2|=n2,.
    If
    n1
    and
    n2
    are co-prime and
    g1
    and
    g2
    commute, then
    |g1g2|=n1n2.

    Sol. First,
    (g1g2)n1n2=(g1n1)n2(g2n2)n1=1.
    On the other hand, suppose
    m
    is a positive integer such that
    (g1g2)m=1.
    Then
    g1m=(g2m)1.
    Hence,
    |g1m|=|g2m|,
    which implies that
    n1/gcd(n1,m)=n2/gcd(n2,m).
    Since
    n1
    and
    n2
    are co-prime, this is possible only if
    gcd(n1,m)=n1
    and
    gcd(n2,m)=n2,
    which implies that
    n1m
    and
    n2m.
    Therefore,
    n1n2=lcm(n1,n2)m
    . Thus,
    n1n2
    is the least positive integer
    m
    such that
    (g1g2)m=1.

  7. Let

    gG with
    |g|=n1n2,
    where
    n1
    and
    n2
    are co-prime positive integers. Then there are elements
    g1,g2G
    such that
    g=g1g2=g2g1
    and
    |g1|=n1,
    |g2|=n2.
    Moreover,
    g1
    and
    g2
    are uniquely determined by these conditions.
    Sol. Since
    n1
    and
    n2
    are co-prime, there exist integers
    a
    and
    b
    such that
    an1+bn2=1.
    Let
    g1=gbn2
    and
    g2=gan1.
    Then,
    g=gan1+bn2=gan1gbn2=g2g1=g1g2,

    since any two powers of

    g commute. Also,
    an1+bn2=1
    implies that
    n1
    and
    b
    are co-prime. Therefore,
    |g1|=n1n2/gcd(bn2,n1n2)=n1n2/n2=n1.
    Similarly,
    |g2|=n2.

    Now, suppose
    h1,h2G
    satisfy
    g=h1h2=h2h1
    and
    |h1|=n1,
    |h2|=n2.
    Then
    gbn2=(h1h2)bn2=h1bn2=h11an1=h1.
    Similarly,
    gan1=h2.
    Hence,
    g1
    and
    g2
    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

    Zp, prove Fermat's theorem: for every integer
    a
    not divisible by
    p,
    ap11modp.

    Sol. The multiplicative group of non-zero elements of
    Zp
    has order
    p1.
    If
    a
    is not divisible by
    p,
    then
    a
    and
    p
    are co-prime, and hence
    abmodp,
    where
    bZp
    with
    b
    and
    p
    co-prime. Hence,
    b
    is in the mutliplicative group, which implies that
    bp1=1.
    Therefore,
    ap11modp.

  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
    Zp,
    prove Wilson's theorem for prime numbers:
    (p1)!1modp.

    Sol. Let
    G={g1,,gn}.
    For each
    giG,
    gi1=gj
    with
    ji
    if and only if
    gi21.
    Therefore, in the product
    g1g2gn,
    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
    Zp,
    namely
    {1,,p1}.
    Then
    (p1)!
    is the product of all elements of
    G.
    If
    nG
    has order
    2,
    then there exists a positive integer
    k<p
    such that
    n2=kp+1kp=n21=(n+1)(n1).
    But
    n1
    and
    k
    are co-prime to
    p,
    and therefore
    n+1
    must be equal to
    p.
    Hence,
    n=p1
    is the only element of
    G
    having order
    2.
    Thus,
    (p1)!1modp.

  10. Let

    HG and let
    g1,g2G.
    Then
    Hg1=Hg2
    if and only if
    g11H=g21H.

    Sol. Recall that
    Hx=Hy
    is equivalent to
    xy1H,
    and that
    xH=yH
    is equivalent to
    x1yH.
    Now,
    Hg1=Hg2g1g21H,
    that is,
    (g11)1(g21)H,
    which is equivalent to
    g11H=g21H.

  11. (i) Let

    JHG. 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
    JHG
    with
    |G:J|=p,
    prime. Then either
    H=J
    or
    H=G.

    Sol. (i) Let
    {xi}iI
    be the distinct representatives of left cosets of
    H
    in
    G,
    and let
    {yr}rR
    be the disinct representatives of
    J
    in
    H.

    Observe that if
    yrJysJ,
    then
    xiyrJxiysJ,
    since
    (xiyr)1(xiys)=yr1ysJ.
    Also, if
    xiH
    and
    xjH
    are distinct, then they are disjoint, which implies that
    xiyrJxjysJ,
    for any
    r,sR.
    Thus,
    {xiyr}iI,rR
    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
    gG,
    g=xia
    for some
    iI,
    and
    aH,
    and then
    h=yrb
    for some
    bJ,
    so
    g=xiyrbxiyrJ.

    Now, since
    {xiyr}iI,rR{xi}iI×{yr}rR
    (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

    HG and
    gG.
    If
    |g|=n
    and
    gmH,
    where
    n
    and
    m
    are co-prime integers, then
    gH.

    Sol. There exist integers
    a
    and
    b
    such that
    am+bn=1.
    Hence,
    gam=g1bn=gH
    (since
    gmH
    ).

Chapter 2

  1. Give an example of a semigroup without an identity element.
    Sol. The set

    N={1,2,} of natural numbers, under addition.

  2. Give an example of an infinite semigroup with an identity element

    e such that no element except
    e
    has an inverse.
    Sol. The set
    N
    under multiplication.

  3. Let

    S be a semigroup and let
    xS.
    Show that
    {x}
    forms a subgroup of
    S
    (of order
    1
    ) if and only if
    x2=x.
    Such an element
    x
    is called an idempotent in
    S.
    (Warning. A semigroup may have several different subgroups of order
    1
    : see 16. Why does a group have only one subgroup of order
    1
    ?)
    Sol. If
    G={x}
    is a subgroup of
    S,
    then
    x2Gx2=x.
    Conversely, if
    x2=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,
    x2=xx=e,
    by cancellation.

  4. 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
    .
    Does
    S
    have an identity element, and if so, what are the units in
    S
    ? Show that every element of
    S
    is an idempotent (15). Deduce that for all
    YS,
    {Y}
    is a subgroup of
    S,
    and that every subgroup of
    S
    has order
    1.

    What happens if
    is replaced by
    ?
    Sol. The intersection of any two subsets of
    X
    is a subset of
    X,
    and
    is associative, hence
    is an associative binary operation on
    S,
    which shows that
    (S,)
    is a semigroup.
    For any
    YS,
    YXYX=Y.
    Hence,
    X
    (which is a subset of itself, and is therefore in
    S
    ) is the identity element of
    S.

    If
    YX,
    then
    YZX
    for all
    ZX,
    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
    YS,
    Y2=YY=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
    is replaced by
    ,
    all the above results hold, except that the identity element becomes
    instead of
    X.

  5. Let

    X be a set with
    |X|=2.
    Do two elements of
    MX
    necessarily commute? Do two elements of
    ΣX
    necessarily commute?
    Note.
    MX
    is the semigroup of all endofunctions of
    X,
    and
    ΣX
    is its group of units.

    Sol. No, not all elements of
    MX
    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=fg=gf.

    ΣX={(),(12)},
    and therefore any two elements of
    ΣX
    commute (since, e.g., every element commutes with itself and with the identity).

  6. Express the permutation

    σ=(12345671574623) as a product of disjoint cycles. Find expressions for
    σ2
    and
    σ1
    as products of disjoint cycles.
    Sol.
    σ=(256)(37),
    σ2=(265),
    and
    σ1=(265)(37).

  7. If

    φ:GH and
    ψ:HJ
    are homomorphisms, then the composite map
    φψ:GJ
    is also a homomorphism.
    Sol. For all
    x,yG,

    (xy)(φψ)=((xy)φ)ψ=((xφ)(yφ))ψ=((xφ)ψ)((yφ)ψ)=x(φψ)y(φψ).

  8. Verify that if

    G is any set of groups then
    is an equivalence relation on
    G.

    Sol. The identity map of any group
    G
    is an isomorphism from
    G
    to itself, hence
    GG.
    If
    GH,
    then there is an isomorphism
    φ:GH,
    and an inverse isomorphism
    ψ:HG,
    which implies that
    HG.
    Finally, if
    GH
    and
    HJ,
    then there exist isomorphisms
    φ:GH
    and
    ψ:HJ.
    Then
    φψ
    is a homomorphism from
    G
    to
    J,
    and since the composition of bijections is a bijection,
    φψ:GJ
    is an isomorphism. Hence,
    GJ.

  9. (i) Calculate the products

    (12)(13)(14) and
    (12)(13)(14)(15)
    in the group
    Σn,
    where
    n
    is an integer with
    n5.

    (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
    (123n)
    can be expressed as a product of
    n1
    transpositions.
    (iii) For any integer
    n>1,
    prove that every non-trivial element of
    Σn
    can be expressed as a product of at most
    n1
    transpositions.
    Sol. (i)
    (12)(13)(14)=(1234),
    and
    (12)(13)(14)(15)=(1234)(15)=(12345).

    (ii)
    (123n)=(12)(13)(1n),
    by induction on
    n.

    (iii) We know that any non-trivial element of
    Σn
    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
    k1
    transpositions. Thus, any non-trivial element of
    Σn
    can be expressed as product of
    ncn1
    transpositions, where
    c
    is the number of cycles in its cycle decomposition.

  10. Let

    n be a positive integer and let
    σΣn.
    Suppose that
    σ
    can be expressed as a product of
    s
    disjoint cycles of lengths
    n1,n2,,ns
    respectively, where
    s,n1,,ns
    are positive integers such that
    n1++ns=n.
    Then
    |σ|
    is the least common multiple of
    n1,n2,,ns.

    Sol. If a permutation is a cycle of length
    k,
    then its order is
    k.
    Let
    σ1,,σs
    be the cycles in the cycle decomposition of
    σ.
    Since disjoint cycles commute,
    σm=σ1mσsm.
    Moreover, if
    σi
    and
    σj
    are disjoint, then so are
    σim
    and
    σjm,
    and hence,
    σm=1σim=1
    for each
    i=1,,s.
    Thus,
    m
    is a multiple of
    ni=|σi|
    for all
    i=1,,s.
    Then
    |σ|,
    being the least such positive integer, is the least common multiple of
    n1,,ns.

  11. Let

    n and
    m
    be positive integers such that
    m
    divides
    n.
    Let
    σ
    be a cycle of length
    n
    in
    Σn.
    Then
    σm
    is the product of
    m
    disjoint cycles of length
    n/m.

    Sol. Let
    n=mk,
    and without loss of generality, let
    σ=(12n).
    Now,
    σm(1)=m+1,
    σm(2)=2m+1,
    ,
    σm((k1)m+1)=1.
    Hence,
    (1,m+1,,(k1)m+1)
    =σ1
    (say) is a cycle of
    σm.
    Next,
    σm(2)=m+2,
    ,
    σm((k1)m+2)=2,
    so
    (2,m+2,,(k1)m+2)=σ2
    is a cycle of
    σm.
    Proceeding similarly,
    (m,2m,,km)=σm
    is a cycle of
    σm,
    and
    km=n.
    Thus the cycle decomposition of
    σm
    is
    σ1σ2σm,
    with each cycle
    σi
    having length
    k=n/m.

  12. Any group which can be embedded in an Abelian group is Abelian.
    Sol. Let

    φ:GH be an embedding of
    G
    into an Abelian group
    H.
    Then for any
    x,yG,
    (xy)φ=(xφ)(yφ)=(yφ)(xφ)=(yx)φ,
    and since
    φ
    is injective, this implies that
    xy=yx.
    Therefore,
    G
    is Abelian.

  13. Prove that if

    X is a non-empty set and
    Y
    a non-empty subset of
    X
    then
    ΣY
    can be embedded in
    ΣX.
    (In particular, whenever
    m
    and
    n
    are positive integers such that
    mn
    then
    Σm
    can be embedded in
    Σn.
    ) Hence or otherwise show that if
    |X|>2
    then
    ΣX
    is non-Abelian.
    Sol. Define a homomorphism
    φ:ΣYΣX
    as follows: For each
    σΣY,
    let
    σφ
    be the permutation of
    X
    defined by
    (x)(σφ)={xσ,xYx,xY.

    To see that

    φ is a homomorphism, first note that
    x(σφ)YxY.
    Now, let
    σ1,σ2ΣY,
    and let
    xX.

    If
    xY,
    then
    x((σ1σ2)φ)=x(σ1σ2)=(xσ1)σ2=(x(σ1φ))(σ2φ).

    If
    xY,
    then
    x(σ1φ)=x(x(σ1φ))(σ2φ)=x=x((σ1σ2)φ).

    Thus,
    (σ1σ2)φ=(σ1φ)(σ2φ).

    Moreover,
    φ
    is injective: for any
    σ1,σ2ΣY,
    and
    yY,
    if
    y(σ1φ)=yσ1=yσ2=y(σ2φ),
    then
    σ1=σ2.
    Therefore,
    φ
    is an embedding of
    ΣY
    in
    ΣX.

    Now, if
    |X|>2,
    then let
    YX
    have three elements. Then
    ΣYΣ3
    is non-Abelian, and since
    ΣY
    embeds in
    ΣX,
    the latter is non-Abelian too (see 24).

  14. If

    φ:GH is a homomorphism and
    G
    is Abelian then
    Imφ
    is Abelian.
    Sol. Let
    xφ,yφImφ.
    Then
    (xφ)(yφ)=(xy)φ=(yx)φ=(yφ)(xφ).

  15. Suppose that

    G1 and
    G2
    are isomorphic groups, and let
    φ:G1G2
    be an isomorphism. If
    K1G1
    and
    K2=K1φ
    then
    |G1:K1|=|G2:K2|.

    Sol. Let
    {xi}iI
    be a system of distinct representatives of left cosets of
    K1
    in
    G1.
    For each left coset
    xiK1,
    (xiK1)φ=(xiφ)K2
    is a left coset of
    K2
    in
    G2.
    If
    (xiφ)K2=(xjφ)K2,
    then
    (xi1xj)φ=(xiφ)1(xjφ)K2=K1φ,
    and therefore
    x1xjK1
    (since
    φ
    is injective), which implies that
    xiK1=xjK2.
    Thus,
    φ
    maps distinct left cosets of
    K1
    to distinct left cosets of
    K2.
    Since
    φ
    is surjective, any element of any coset of
    K2
    can be written as
    xφ,
    for some
    xG1,
    and then
    xxiK1,
    for some
    iI.
    Thus,
    xφ(xiφ)K2,
    which shows that every left coset of
    K2
    in
    G2
    is of the form
    (xiφ)K2,
    for some
    iI.
    Hence, the image of
    {xi}iI
    under
    φ
    is a system of distinct representatives of left cosets of
    K2
    in
    G2,
    and is in bijection with
    {xi}iI.
    Therefore,
    |G1:K1|=|G2:K2|.

  16. 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=Σ3,
    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 24).

  17. (i) Any infinite cyclic group is isomorphic to

    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=g
    be an infinite cyclic group. Define
    φ:GZ+
    by
    (gn)φ=n,
    for each
    gnG.
    Note that
    gn=gm
    if and only if
    gnm=1,
    which is equivalent to
    n=m,
    since
    g
    has infinite order. Thus,
    φ
    is well-defined and injective. Finally, for any
    nZ+,
    (gn)φ=n,
    hence
    φ
    is surjective. Thus,
    φ
    is an isomorphism from
    G
    to
    Z+.

    (ii) Let
    gG
    be non-trivial. Then
    g,
    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
    gm
    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|
    (
    1
    ) is prime.

  18. (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<HG,
    JH
    then every element of
    G
    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
    gG
    having infinite order. Let
    H=g
    and
    J=g2.
    Then
    J<H,
    but
    JZ+H,
    which contradicts the given condition.

  19. Show that

    Z+ has infinitely many distinct subgroups. Deduce that every infinite group has infinitely many distinct subgroups.
    Sol. For each
    nN,
    we have a subgroup
    nZ+.
    For any integer
    m,
    mn
    if and only if
    m=nk,
    for some integer
    kZ,
    i.e.,
    nm.
    Thus,
    n=m
    only if
    mn
    and
    nm,
    which implies that
    m=n
    (since
    m,nN
    ). Thus,
    {nnN}
    is a collection of infinitely many distinct subgroups of
    Z+.

    Now, let
    G
    be any infinite group. If
    G
    contains an element, say
    g,
    of infinite order, then
    gZ+
    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
    xG,
    then, since
    x
    is finite, there are infinitely many
    yG
    such that
    xy.
    Thus,
    G
    has infinitely many distinct subgroups.

  20. No two of the groups

    Z+,
    Q+,
    R+
    are isomorphic. (Remark. It is in fact true that
    R+C+.
    This can be proved by regarding
    R
    and
    C
    as vector spaces over
    Q,
    and using vector space theory and facts about infinite cardinals.)
    Sol. Let
    φ:Z+Q+
    be any injective homomorphism, and let
    r=1φQ+.
    Then,
    r/2Q+,
    but there exists no
    nZ+
    such that
    r/2=nφ,
    since otherwise
    1φ=r=2(r/2)=2(nφ)=(2n)φ
    would imply that
    2n=1
    (by injectivity of
    φ
    ), but
    Z+
    contains no such
    n.
    Therefore,
    φ
    is not surjective. This shows that there is no isomorphism from
    Z+
    to
    Q+.
    Since
    Q+R+,
    it follows that
    Z+
    cannot be isomorphic to
    R+
    either.
    Next, suppose that
    ψ:Q+R+
    is an isomorphism. For any
    m/nQ+,
    (m/n)ψ=yR+
    implies that
    mψ=n((m/n)ψ)=ny.
    It follows that if
    1ψ=xR+,
    then
    (m/n)ψ=mx/nR+.
    Now, let
    m/nQ+
    be such that
    2=(m/n)ψ=mx/n,
    and let
    p/qQ+
    be such that
    2=(p/q)ψ=px/n.
    Then
    2=2/2=(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
    Q+
    is countable and
    R+
    is uncountable, and hence there is no bijection between them, let alone an isomorphism of groups.

  21. 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
    φ,ψHom(G,A),

    φ+ψ:GA
    is defined by
    φ+ψ:ggφgψ
    for all
    gG.
    Verify that
    φ+ψHom(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(Z+,A)A.

    Sol. Let
    φ,ψHom(G,A).
    Then, for any two elements
    g,hG,

    (gh)(φ+ψ)=(gh)φ(gh)ψ=gφhφgψhψ=gφgψhφhψ[A is Abelian]=g(φ+ψ)h(φ+ψ).

    Thus,

    φ+ψHom(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
    φHom(G,A),
    the map
    ψ:GA
    defined by
    gψ=(gφ)1,
    for all
    gG,
    is a homomorphism from
    G
    to
    A
    and is an inverse with respect to
    +,
    as shown below:
    g(φ+ψ)=gφgψ=gφ(gφ)1=1

    which implies that

    φ+ψ 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
    aA,
    let
    φa:Z+A
    be the function defined by
    nφ=an,
    for all
    nZ+.
    Then
    φa
    is a homomorphism from
    Z+
    to
    A,
    since
    (m+n)φ=am+n=aman=(mφ)(nφ).

    Next, define a map

    f:AHom(Z+,A) by
    af=φa,
    for all
    aA.

    First, observe that
    f
    is a homomorphism: if
    a,bA,
    (ab)f=φab,
    and
    af+bf=φa+φb.
    To see that these two maps are equal, let
    nZ+.
    Then
    nφab=(ab)n=anbn=nφanφb=n(φa+φb).

    Next, to show that

    f is bijective, observe that every element
    φ
    of
    Hom(Z+,A)
    is uniquely determined by the value of
    1φ
    : if
    φHom(Z+,A),
    then
    nφ=n(1φ),
    which shows that every
    φHom(Z+,A)
    is of the form
    φa,
    with
    a=1φ,
    and that
    φa=φba=b.
    Therefore,
    f
    is an isomorphism from
    A
    to
    Hom(Z+,A).

  22. Show that if

    φ:GH is a homomorphism,
    gG,
    and
    n
    is a positive integer such that
    gn=1,
    then
    (gφ)n=1.
    Hence, by considering the elements
    z
    of
    C×
    satisfying
    z3=1,
    or otherwise, show that
    C×
    is not isomorphic to either
    Q×
    or
    R×.

    Sol.
    gn=1(gn)φ=1φ(gφ)n=1.

    Now, let
    φ:C×G
    be any homomorphism, where
    G=Q×
    or
    R×.
    Let
    z=e2πi/3C×.
    Then
    z3=1(zφ)3=1.
    But the only such element in
    G
    is
    1,
    hence
    zφ=1=1φ,
    which shows that
    φ
    is not injective, and hence cannot be an isomorphism.

  23. Prove that

    Q× is not isomorphic to
    R×.
    (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=bn.
    )
    Sol. Suppose, to the contrary, that there exists an isomorphism
    φ:R×Q×.
    Then,
    (1)2=1
    ((1)φ)2=1φ=1
    (1)φ=1,
    since
    1
    and
    1
    are the only square roots of
    1
    in
    Q,
    and
    φ
    is injective. This implies that
    (x)φ=(xφ)
    for all
    xR×.

    Now, let
    x=2φ1R×.
    If
    x>0,
    then
    x=(x)2
    2=xφ=((x)φ)2,
    which is impossible, as
    2Q.
    If
    x<0,
    then
    x=(x)2
    2=(x)φ=((x)φ)2,
    which is impossible, as
    2Q.
    Thus, there is no such isomorphism
    φ.

  24. Prove that there is no field

    F such that
    F+F×.
    (Hint. Assume that there is a field
    F
    with an isomorphism
    φ:F×F+
    and consider
    (1)φ.
    )
    Sol. If possible, let
    φ:F×F+
    be an isomorphism, and let
    v=(1)φF+.
    Then,
    (1)2=1
    2v=2((1)φ)=1φ=0.
    This implies that
    v=0
    (for, if
    v0,
    then
    2v=0
    2×1=0
    1=1
    v=1φ=0
    ), and therefore,
    1=1.
    Now, for all
    xF×,
    (x1)φ=(xφ)=xφ
    x1=x
    x2=1.
    But this implies
    x21=0
    (x+1)(x1)=0
    (x1)2=0
    x=1.
    Hence,
    F={0,1},
    and therefore
    F×={1}
    cannot be isomorphic to
    F+={0,1}.

  25. Let

    G be a field. Define a binary operation
    on
    F
    by
    ab=a+babfor all a,bF.
    Prove that the set of all elements of
    F
    distinct from
    1
    forms a group
    F
    with respect to the operation
    ,
    and that
    FF×.

    Sol. Let
    F={aFa1}.

    Observe that if
    x,yF,
    then
    (1x)(1y)=1xy+xy.
    Therefore, if
    a,bF,
    then
    x=1a
    and
    y=1b
    are in
    F×,
    and
    ab=1xy.
    From this, it is easy to see that
    is an assocative and commutative binary operation on
    F,
    and that
    11=0
    is the identity element and for each
    aF,
    ab=0
    if
    1(1a)(1b)=0,
    i.e.,
    b=aa1
    (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
    φ:F×F
    by
    xφ=1x,
    for all
    xF×.
    This is well defined, since
    xF×
    x0
    1x1.
    It is also obvious that
    φ
    is bijective, with inverse
    aφ1=1a,
    for all
    aF.
    The earlier observation shows that
    (xy)f=(xf)(yf),
    hence
    f
    is an isomorphism.

  26. The set of all positive rational numbers forms a subgroup

    Qpos+ of
    Q×,
    and there is a surjective homomorphism
    |  |:Q×Qpos×
    which is not an isomorphism.
    Is
    Qpos×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
    Q×
    to
    Qpos×.
    Finally,
    1,1Q×,
    and
    |1|=|1|=1,
    hence
    |  |
    is not an isomorphism.
    To see that
    Qpos×Q+,
    suppose, to the contrary, that there exists an isomorphism
    φ:Qpos×Q+.
    Let
    a=2φ.
    Then
    a/2Qpos×
    a/2=xφ,
    for some
    xQpos×.
    But
    2φ=a=2(a/2)=2(xφ)=(x2)φ
    x2=2,
    which is impossible, since
    2Qpos×.
    Hence,
    Qpos×Q+.

  27. For any positive integer

    n, the only homomorphism
    φ:CnC+
    is the trivial homomorphism.
    Sol. Let
    φ:CnC×
    be any homomorphism, and let
    z=(e2iπ/n)φ.
    Then,
    (e2iπ/n)n=1
    nz=1φ=0
    z=0.
    Since every element of
    Cn
    is a power of
    e2iπ/n,
    this implies that
    φ
    maps all elements of
    Cn
    to
    0,
    and is therefore the trivial homomorphism.

  28. Let

    n be a positive integer. Then
    (i)
    Zn+Cn.

    (ii)
    Zn×
    is an Abelian group of order
    φ(n),
    where
    φ
    is Euler's function (see 5).
    (iii) By considering orders of elements in
    Zn×,
    prove the Euler-Fermat theorem:
    mφ(n)1modn,

    whenever

    m is an integer co-prime to
    n.
    (This generalizes 8).

    Sol. (i) Let
    ω=e2iπ/n,
    and define
    φ:Zn+Cn
    by
    kφ=ωk,
    for all
    kZn+.
    Then for
    k,lZn+,
    (k+l)φ=ωk+l,
    since
    ωt=ωs
    whenever
    tsmodn.
    Thus,
    (k+n)φ=(ωk)(ωl)=(kφ)(lφ).
    Since
    |Zn+|=|Cn|
    and
    φ
    is clearly surjective,
    φ
    is an isomorphism.
    (ii) Multiplication is associative and commutative in
    Zn×,
    1Zn×,
    and all elements of
    Zn×
    are invertible, by definition. If
    a,bZn×,
    then
    (ab)(b1a1)=1,
    hence
    abZn×.
    Therefore,
    Zn×
    is an Abelian group.
    Now, let
    mZn.

    First, suppose that
    m
    is not co-prime to
    n,
    so
    gcd(m,n)=d1.
    Then
    n/d
    is a non-zero element of
    Zn,
    and
    m(n/d)=n(m/d)=0
    (since
    dm
    ). Thus,
    m
    is a zero-divisor and hence is not a unit of
    Zn.

    Next, suppose that
    m
    is co-prime to
    n.
    Then there exist integers
    a
    and
    b
    such that
    am+bn=1
    am1modn.
    Thus,
    a
    (modulo
    n
    ) is a multiplicative inverse of
    m,
    and
    m
    is a unit of
    Zn.
    Therefore,
    Zn×
    consists of all the positive integers less than
    n
    and co-prime to it. Hence,
    |Zn×|=φ(n).

    (iii) From (ii), if
    m
    is co-prime to
    n,
    then
    mamodn
    for some
    aZn×.
    Then
    mφ(n)aφ(n)1modn,
    since
    φ(n)=|Zn×|.

  29. (i) Let

    G be a group and
    n
    a positive integer such that
    gn=1
    for all
    gG.
    Show that if
    φ:GC×
    is a homomorphism then
    ImφCn.
    Hence show that
    Hom(G,C×)Hom(G,Cn)
    (cf. 33).
    (ii) Deduce that if
    G
    is a finite group then so is
    Hom(G,C×).
    (Remark. If
    G
    is a finite Abelian group then in fact
    Hom(G,C×)G,
    but we do not yet have the means of proving this. See 454.)
    Sol. (i) For each
    gG,
    gn=1
    (gφ)n=1
    gφCn.

    Now, let
    f:Hom(G,Cn)Hom(G,C×)
    be the inclusion map, which is always an injective homomorphism. As proved above,
    f
    is surjective as well. Hence,
    Hom(G,C×)Hom(G,Cn).

    (ii) If
    G
    is finite, of order
    n,
    then
    gn=1
    for all
    gG.
    Hence, by (i),
    Hom(G,C×)Hom(G,Cn),
    which is finite since
    G
    and
    Cn
    are both finite.

  30. Let

    G be a finite group such that
    g2=1
    for all
    gG.
    Show that
    GV+
    for some finite dimensional vector space
    V
    over
    Z2.
    Deduce that
    |G|=2m
    for some integer
    m0.
    (Hint. By 3,
    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
    +:V×VV
    by
    g+h=gh,
    for all
    g,hV.
    Then
    (V,+)
    is clearly an Abelian group. Define
    :Z2×VV
    by
    ng=gn,
    for all
    nZ2
    and
    gV.

    Now, for
    a,bZ2
    and
    g,hV,

    (a+b)(g+h)=(gh)a+b=gahagbhb=ag+ah+bg+bh,

    a(bg)=(gb)a=gab=abg,
    and
    1g=g1=g.

    Thus,
    (V,+,)
    is a vector space over
    Z2.

    Let
    φ:GV
    be the function defined by
    gφ=g,
    for all
    gG.
    From the definition of
    +,
    it is evident that
    (gh)φ=gφ+hφ,
    for all
    g,hG.
    Since
    V=G,
    φ
    is obviously a bijection. Hence.
    φ
    is an isomorphim from
    G
    to
    V+.
    Now,
    V
    is a finite dimensional vector space over
    Z2
    (since it is finite), and therefore is isomorphic to
    Z2m,
    for some integer
    m0.
    Hence,
    |G|=|V|=|Z2|m=2m.

  31. Let

    F be a field and let
    m,n
    be positive integers with
    mn.
    Then
    (i)
    GLm(F)
    can be embedded in
    GLn(F),

    (ii)
    GLn(F)
    is non-Abelian for all
    n2.

    Sol. (i) Let
    k=nm.
    Define a map
    φ:GLm(F)GLn(F)
    by
    Aφ=(A0m,k0k,mIk)

    where

    0m,k and
    0k,m
    are zero matrices of orders
    k×m
    and
    m×k
    respectively, and
    Ik
    is the identity matrix of order
    k.

    From the laws of multiplication of partitioned matrices, it is clear that
    φ
    is a homomorphism. Since equality of two
    n×n
    matrices implies equality of all corresponding pairs of submatrices,
    φ
    is an embedding.
    (ii) Let
    A=(1101),
    and
    B=AT.
    Then
    A,BGL2(F)
    and
    ABBA,
    which implies that
    GL2(F)
    is non-Abelian. From (i),
    GL2(F)
    can be embedded in
    GLn(F)
    for all
    n2,
    and therefore,
    GLn(F)
    is non-Abelian as well (see 24).

  32. GL2(Z2)Σ3.
    Sol. Consider the vector space
    V=Z22
    with the standard basis
    B={u=[10],v=[01]}.
    Then
    GL2(Z2)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)ΣX.
    This further implies that
    GL2(Z2)Σ3.

  33. Let

    G be the set of all matrices of the form
    (ab0c),
    where
    a,
    b,
    c
    are real numbers such that
    ac0.
    Prove that
    G
    forms a subgroup of
    GL2(R),
    and that the set
    H
    of all elements of
    G
    in which
    a=c=1
    forms a subgroup of
    G
    isomorphic to
    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
    ac0,
    hence
    GGL2(R),
    and
    G
    is clearly non-empty. Now, let
    A=(a1b10c1)
    and
    B=(a2b20c2)
    be any two elements of
    G.
    Then
    (45.1)AB1=(a1b10c1)×1a2c2(c2b20a2)=1a2c2(a1c2a2b1a1b20a2c1),

    hence

    AB1G. Therefore,
    GGL2(R).

    Let
    H
    be the set of all the matrices in
    G
    whose diagonal elements are both
    1.
    From
    (45.1),
    if
    A,BH,
    then
    AB=A(B1)1=(1b1+b201).
    Define
    φ:R+G
    as the function that maps each
    bR+
    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
    Imφ=H.
    Hence,
    HG
    and
    HR+.

    To find all elements of
    G
    of order
    2,
    let
    A=(ab0c).
    Then
    A2=(a2b(a+c)0c2).
    Therefore,
    A2=I
    if and only if
    |a|=|c|=1
    and either
    b=0
    or
    c=a.

    Let
    A=(1001)
    and
    B=(1101).
    Then
    A
    and
    B
    are of order
    2,
    but
    AB=(1101)
    is a non-identity element of
    HR+,
    and hence has infinite order.

  34. (i) If

    R is a ring with a multiplicative identity
    1
    then
    R×
    can be embedded in
    AutR+.
    (Hint. See 2.11.)
    (ii)
    AutZ+Z×
    and
    AutZn+Zn×
    for every positive integer
    n.

    Sol. (i) For each
    uR×,
    define a function
    αu:R+R+
    by
    xαu=xu,
    for all
    xR+.
    Distributivity implies that
    αu
    is an endomorphism of
    R+.
    Since
    u
    is a unit, it has a multiplicative inverse
    u1R×,
    and for all
    xR+,
    (xαu)αu1=(xu)u1=x,
    which implies that
    αuαu1
    is the identity automorphism. Since
    u1R+
    and
    (u1)1=u,
    this also implies that
    αu1αu
    is the identity automorphism. Thus,
    αu
    is an automorphism of
    R+.

    Now, define a map
    φ:R×AutR+
    by
    uφ=αu,
    for all
    uR×.
    This is well-defined, as shown above. If
    u,vR×,
    then
    xαuαv=xuv=xαuv,
    for all
    xR+.
    That is,
    (uφ)(vφ)=(uv)φ,
    and hence
    φ
    is a group homomorphism. If
    uφ=vφ,
    then
    αu=αv
    u=1u=1αu=1αv=1v=v.
    Hence,
    φ
    is an embedding.
    (ii) Let
    R
    be the ring
    Z
    or
    Zn.
    Let
    φ
    be the embedding of
    R×
    into
    AutR+
    defined in (i). Let
    α
    be any automorphism of
    R+,
    and let
    m=1α.
    Note that
    m=1αmR×.
    Since
    R+=1,
    every element of
    nR+
    can be written in the form
    n1
    , and hence
    (n1)α=n(1α)=nm=nαm.
    But
    αm=mφ.
    This shows that
    φ
    is surjective, and hence bijective. Therefore,
    AutR+R×,
    for
    R=Z
    or
    Zn.

    Note. To see that
    mR×,
    observe that, since
    α
    is an automorphism,
    m(1α1)=(m1)α1=mα1=1,
    which shows that
    m1=1α1,
    and hence
    mR×.

  35. Let

    V be a vector space
    0
    over a field
    F.
    Then
    (i)
    GL(V)AutV+.
    (See 2.15, 2.16.)
    (ii) If
    F=Zp
    and
    V
    has finite dimension
    n
    over
    Zp
    then
    AutV+GLn(Zp).

    Sol. (i) If
    αGL(V),
    then for each
    u,vV,
    (u+v)α=uα+vα,
    by linearity, and
    α
    is invertible by definition, hence it is an automorphism of
    V+.

    (ii) Suppose
    V
    has finite dimension over
    F=Zp.

    Let
    αAutV+.
    Then for any
    u,vV,
    (u+v)α=uα+vα
    , which implies that for any
    a,bZp,
    considering
    a
    and
    b
    as integers, we have
    (au+bv)α=a(uα)+b(vα).
    Hence,
    αGL(V).
    From (i), this implies that
    AutV+=GL(V),
    and we know that
    GL(V)GLn(Zp).

  36. If

    G1G2 then
    AutG1AutG2
    and
    InnG1InnG2.

    Sol. Let
    φ:G1G2
    be an isomorphism.
    Given an automorphism
    α1
    of
    G1
    , define
    α2:G2G2
    as
    α2=φ1α1φ.
    Then
    α2,
    being a composition of isomorphisms, is itself an isomorphism, and hence is an automorphism of
    G2.

    Now, define
    f:AutG1AutG2
    as
    α1f=α2,
    for each
    α1AutG1,
    . Then
    f
    is an isomorphism: For each
    α1,β1AutG1,

    (α1β1)f=φ1(α1β1)φ=(φ1α1φ)(φ1β1φ)=(α1f)(β1f)

    and the map

    g:AutG2AutG1 defined by
    α2g=φα2φ1,
    for all
    α2AutG2,
    is an inverse of
    f.

    Hence,
    AutG1AutG2.

    Now consider the restriction of
    f
    to
    InnG1.
    Let
    α1InnG1,
    and
    α2=α1f.
    Thus, there exists
    gG,
    such that for each
    xG1,
    xα1=g1xg
    . Let
    h=gφ,
    and observe that
    (xα1)φ=(gφ)1(xφ)(gφ)=h1(xφ)h.
    From this it follows that for any
    yG2,
    yα2=h1yh.
    Thus,
    α2InnG2.
    Also, if
    α2
    is any inner automorphism of
    G2,
    with say,
    yα2=h1yh,
    for some
    hG2,
    for all
    yG2,
    then clearly,
    α2=α1f,
    where
    α1
    is defined by
    xα1=g1xg,
    for all
    xG1,
    with
    g=hφ1.
    Thus, the restriction of
    f
    to
    InnG1
    is surjective onto
    InnG2,
    and is injective as well, since
    f
    is injective. Hence, this map is an isomorphism from
    InnG1
    to
    InnG2.

  37. Define a relation

    on
    G
    by
    xyif and only ifg1xg=yfor some gG.
    Show that this relation
    of conjugacy is an equivalence relation on
    G.

    Sol. Let
    x,y,zG.

    Since
    11x1=x,
    xx.

    If
    xy,
    then
    g1xg=y,
    for some
    gG.
    This implies that
    (g1)1y(g1)=x,
    and hence
    yx.

    If
    xy
    and
    yz
    , then
    g1xg=y
    and
    h1yh=z,
    for some
    g,hG.
    Conjugating the former equation by
    h,
    h1g1xgh=h1yh
    (gh)1x(gh)=z,
    and hence
    xz.

    Thus,
    is reflexive, symmetric, and transitive.

  38. If

    αAutG and
    xG
    then
    |xα|=|x|.
    In particular, conjugate elements of a group have the same order.
    Sol. If
    xn=1
    , then
    (xα)n=(xn)α=1α=1.
    This implies that
    |xα|
    divides
    |x|.
    But
    αAutG
    α1AutG
    and
    x=(xα)α1,
    hence
    |x|
    divides
    |xα|
    as well. Thus,
    |x|=|xα|.

  39. Find a group

    G with a subgroup
    K
    and element
    g
    such that
    g1KgK.

    Sol. Let
    G=Σ3,
    K={(),(12)},
    and
    g=(13).

    Now,
    g1(12)g=(13)(12)(13)=(23).
    Hence,
    g1Kg={(),(23)}K.

  40. (i) The map defined by

    gg1 for all
    gG
    is an automorphism of
    G
    if and only if
    G
    is Abelian.
    (ii) If
    G
    is any group for which
    g21
    for some
    gG
    then
    AutG1.
    (Remark. In fact
    AutG1
    if and only if
    |G|>2.
    The proof is completed by observing that if
    g2=1
    for all
    gG,
    so that by 3
    G
    is Abelian, then
    G
    has the structure of a vector space
    V
    over
    Z2,
    and any invertible linear map
    VV
    is an automorphism of
    G.
    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
    x,yG,
    (xy)1=x1y1.
    Equivalently (taking inverse on both sides),
    xy=(x1y1)1=yx,
    that is,
    G
    is Abelian.
    (ii) If
    G
    is Abelian, then the map
    xx1
    is an automorphism, and it is non-trivial, since
    g21
    gg1.
    If
    G
    is non-Abelian, then let
    x
    and
    y
    be elements of
    G
    that do not commute. Then the inner automorphism
    zx1zx
    is a non-trivial automorphism, since
    yx1yx.

    Thus, in either case,
    AutG1.

  41. (i) Let

    αAutG and let
    H={gGgα=g}.
    Prove that
    H
    is a subgroup of
    G:
    it is called the fixed point subgroup of
    G
    under
    α.

    (ii) Let
    n
    be a positive integer and
    F
    a field. For any
    n×n
    matrix
    y
    with entries in
    F,
    let
    y
    denote the transpose of
    y.
    Show that the map
     :GLn(F)GLn(F),
    defined by
     :x(x1)
    for all
    xGLn(F),
    is an automorphism of
    GLn(F),
    and that the corresponding fixed point subgroup consists of all orthogonal
    n×n
    matrices with entries in
    F
    (that is, matrices
    y
    such that
    yy=1
    ).
    Prove (by assuming the contrary and considering determinants) that
    is an outer automorphism of
    GLn(F)
    if
    FZ2
    and
    FZ3.
    (Remark. In fact,
    is an outer automorphism of
    GLn(F)
    unless either
    F=Z2
    and
    n2
    or
    F=Z3
    and
    n=1
    ).

    Sol. (i)
    H,
    since
    1α=1
    1H.
    If
    x,yH,
    then
    xα=x
    and
    yα=y
    (xy1)α=xα(yα)1=xy1
    xy1H.
    Thus,
    HG.

    (ii) Since both transposition and inversion preserve invertibility and are self-inverse maps, their composition,
    , is a bijection from
    GLn(F)
    to itself. To see that it is a homomorphism, observe that for all
    x,yGLn(F),

    (xy)=((xy)1)=(y1x1)=(x1)(y1)=xy.

    Thus,

    AutGLn(F).
    Now, let
    xGLn(F).
    Then
    x
    is a fixed point of
    if and only if
    x=x=(x1)=(x)1,
    which is equivalent to
    xx=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
    GLn(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
    gGLn(F)
    such that for all
    xGLn(F),
    x=g1xg.
    Then
    detx=det(g1xg)=(detg)1(detx)(detg)=detx.

    But

    detx=det((x1))=(detx)1. If
    FZ2
    and
    FZ3,
    then
    F
    has a non-zero element
    k
    such that
    aa1,
    and there exists
    xGLn(F)
    with
    detx=a
    (for example, the diagonal matrix with
    x11=a
    and
    xii=1
    for all
    i>1
    ). Thus, for this matrix
    x,
    detx(detx)1,
    which is a contradiction. Therefore,
    is not an inner automorphism.

  42. Let

    αAutG. Then
    α
    is said to be fixed-point-free if the fixed point subgroup of
    G
    under
    α
    is trivial (see 53); that is, if
    gαg
    whenever
    1gG.

    (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
    α
    is a fixed-point free automorphism of the finite group
    G.
    Show that
    {gαg1gG}=G.
    Deduce that if
    |α|=2
    then, for all
    xG,

    xα=x1,
    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
    α:xx1,
    defined for all
    xG,
    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
    gG,
    by
    ggαg1,
    is injective. Then use 52 (i) and 1.13.)
    Sol. (i) Let
    φ:GG
    be the map
    gφ=gαg1,
    for all
    gG.
    Then
    φ
    is injective: if
    gφ=hφ,
    then
    gαg1=hαh1
    (h1g)α=h1g
    h1g=1,
    since
    α
    is fixed-point-free, and therefore
    g=h.
    Since
    G
    is finite,
    φ
    is bijective, and therefore
    G=Imφ={gαg1gG}.

    Now, let
    xG.
    Then
    x=gαg1,
    for some
    gG,
    which implies that
    xα=gα2(g1)α=g(gα)1=(gαg1)1=x1.

    Now, from 52,

    G must be Abelian, since
    α
    is an automorphism of
    G.
    Also,
    |G|>1,
    for otherwise
    α
    would be trivial, whereas we know that
    |α|=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
    α
    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 52, we know that if
    G
    is Abelian, then
    αAutG.
    To see that
    α
    is fixed-point-free, let
    x
    be a fixed point of
    α.
    Then
    xα=x1=x
    x2=1.
    Thus, either
    x=1
    or
    |x|=2.
    But since
    |G|
    is odd,
    |x|
    cannot be even, hence the only fixed point of
    α
    is
    1.

  43. Let

    X be any non-empty set. Let
    c
    be any positive real number and define, for all
    x,yX,

    d(x,y)={0if x=ycif xy.
    Show that
    d
    is a distance function for
    X,
    and that for this metric space
    IsomX=ΣX.

    Sol. By definition,

    d(x,y)=0 if and only if
    x=y,
    and
    d(x,y)=d(y,x),
    for all
    x,yX.
    Now, let
    x,y,zX.
    If
    x=z,
    then
    0=d(x,z)d(x,y)+d(y,z).
    If
    xz,
    then
    y
    is distinct from at least one of
    x
    and
    z,
    and hence
    c=d(x,z)d(x,y)+d(y,z).
    Therefore,
    d
    is a distance function on
    X
    .
    Now, let
    σΣX.
    Then for any
    x,yX,
    xyxσyσ,
    which implies that
    d(x,y)=cd(xσ,yσ)=c,
    and
    d(x,y)=0d(xσ,yσ)=0.
    Hence,
    d(xσ,yσ)=d(x,y).
    Therefore,
    σIsomX.

  44. View

    R as the Euclidean line
    E1
    with the usual distance function (that is, for
    x,yR,
    d(x,y)=|xy|
    ).
    (i) For each
    aR,
    the map
    τa:xx+afor all xR
    is an isometry of
    R,
    called a translation, and
    R+T={τaaR}IsomR.
    (ii) For each
    aR,
    the map
    εa:xaxfor all xR
    is an isometry of
    R
    called a reflexion, and
    εa2=1(=τ0)andεaτb=τbεafor all a,bR.
    (iii) Every isometry of
    R
    is either a translation or a reflexion.
    (iv)
    IsomR
    is a non-Abelian group and
    T
    is an Abelian subgroup of index
    2.

    Sol. (i) For all

    aR,
    d(xτa,yτa)
    =
    d(x+a,y+a)
    =
    |(x+a)(y+a)|
    =
    |xy|
    =
    d(x,y)
    . Thus,
    τa
    is an isometry of
    R
    . Define
    τ:RIsomR
    by
    aτa
    . If
    a,bR
    , then
    τ(a+b)=τa+b
    is the isometry that maps each
    xR
    to
    (a+b)+x
    =
    (x+a)+b
    =
    (xτa)τb
    . Hence,
    τ(a+b)
    =
    τaτb
    =
    τ(a)τ(b)
    , and therefore
    τ
    is a homomorphism from
    R+
    to
    IsomR
    . Now, if
    a,bR
    are such that
    τ(a)=τ(b)
    , then for
    a
    =
    0+a
    =
    0τa
    =
    0τ(a)
    =
    0τb(0)
    =
    0τb
    =
    0+b=b
    . Thus,
    τ
    is injective. Clearly,
    Imτ={τaaR}
    IsomR
    .
    (ii) For all
    aR
    ,
    d(xεa,yεa)
    =
    d(ax,ay)
    =
    |(ax)(ay)|
    =
    |yx|
    =
    d(x,y)
    . Thus,
    εa
    is an isometry of
    R
    . Since,
    (xεa)εa
    =
    (a(ax))
    =
    x
    ,
    εa2=1
    . Also,
    (xεa)τb
    =
    (ax)τb
    =
    ax+b
    =
    a(xb)
    =
    (x+(b))εa
    =
    (xτb)εa
    . Hence,
    εaτb=τbεa
    .
    (iii) We know that for all
    aR
    ,
    τa,εaIsomR
    . We will show that every isometry of
    R
    is either
    τa
    or
    εa
    for some
    aR
    . Let
    φ
    be any isometry of
    R
    , and let
    a=0φ
    . Now, if
    xR
    ,
    d(x,0)=d(xφ,0φ)
    implies that
    |x|=|xφa|
    , and hence
    xφ=a+x
    or
    xφ=ax
    . That is,
    φ=τa
    or
    φ=εa
    . Therefore, every isometry of
    R
    is either a translation or a reflexion.
    (iv) We know from (ii) that, for example,
    ε1τ1=τ1ε1
    , and
    τ1τ1
    . Hence,
    IsomR
    is non-Abelian. From (i),
    T
    is a group isomorphic to
    R+
    , and hence is an Abelian subgroup of
    IsomR
    . To show that the index of
    T
    in
    IsomR
    is
    2
    , we will show
    T
    and
    ε0T
    are the only two left cosets of
    T
    in
    IsomR
    . We know that every isometry of
    R
    is etiher
    τaT
    or
    εa
    , for some
    aR
    . Hence, every element of
    IsomR
    not in
    T
    is of the form
    εa
    , for some
    aR
    . But
    εa=ε0τaε0T
    .

  45. Let the notation be as in 56. Show that for every

    nZ,
    τ1n=τn,
    and that the elements of the symmetry group
    SR(Z)
    are just the isometries
    τ1nfor all nZandτ1nε0for all nZ.
    Note that
    ε02=1
    and
    ε0τ1=τ11ε0.
    The group
    SR(Z)
    is called the infinite dihedral group and denoted by
    D.

  46. Let

    n be an integer,
    n3.
    Then
    D2n
    can be embedded in
    Σn.
    Moreover,
    D6Σ3
    but, whenever
    n>3
    ,
    D2nΣn.
    (It may be assumed that an isometry of
    E2
    is uniquely determined by its effect on any 3 non-collinear points.)

  47. Let

    n be an integer,
    n3,
    and let
    L=D2n.
    Let
    J=ρ,
    where
    ρ
    is defined as in 2.24. Show that every element of
    LJ
    has order
    2.
    Deduce that
    J
    is the only cyclic subgroup of
    L
    of order
    n.

  48. Every group of order

    6 is isomorphic to either
    C6
    or
    D6.
    Hence, in the notation of chapter 1,
    ν(6)=2.

    (Hints. Let
    G
    be a non-cyclic group of order
    6.
    By 42,
    G
    has an element
    x
    of order
    3.
    Let
    y
    be an element of
    G
    other than
    1,x,x2.
    Then
    G={1,x,x2,y,xy,x2y}
    ,
    y2=1
    and
    yx=x2y.
    )

  49. (i) For any 2 points

    s,
    t
    of
    E2
    there is a unique translation
    τa
    of
    E2
    which maps
    s
    to
    t.

    (ii) For any 2 points
    s,
    t
    of
    E2,
    if
    τa
    is the unique translation of
    E2
    which maps
    s
    to
    t
    then
    τa1Rot(E2;s)τa=Rot(E2;t).
    Thus any 2 groups of rotations are conjugate subgroups of
    IsomE2.