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    Suppose that G is a finite Abelian group and G has no element of order 2. Show that the mapping g-->g^2 is an automorphism of G. Show, by example, that if G is infinite the mapping need not be an automorphism (hint: consider Z).

    © BrainMass Inc. brainmass.com October 7, 2022, 7:21 pm ad1c9bdddf
    https://brainmass.com/math/linear-transformation/finite-abelian-group-359983

    SOLUTION This solution is FREE courtesy of BrainMass!

    It is given that G is a finite Abelian group, and has no element of order 2.
    Let us denote the mapping g - > g^2, by f. That is, define f: G -> G as
    f(g) = g^2, for all elements g of G

    In order to show that f is an automorphism, we must show that f is an isomorphism that is onto (that is, we must show that the f is a homomorphism from that is injective, and also surjective - that f(G), the image of G, is G itself).

    Step 1:
    First, prove that f is a homomorphism:
    Let a and b be two elements of G
    Then f(a)f(b) = (a^2)(b^2)
    = (ab)^2 [Since G is Abelian]
    = f(ab)
    Hence, f is a homomorphism.

    Step 2:
    Prove that f is injective:
    Since G has no element of order 2:
    for any element g in G, g^2 = e implies that g = e (where e is the identity element)
    Therefore, f(g) = e implies that g = e [Since f(g) = g^2]
    Thus the kernel of f, is
    ker f = {e} [Definition: ker f = {g in G | f(g) = e}]
    But the kernel of a homomorphism is trivial (consists only of the identity element) if and only if it is injective.
    Hence, f is injective

    Step 3:
    Prove that f is surjective (onto):
    Since G is finite, and f is injective:
    f is an injective function between two sets of the same cardinality (the "two" sets here being G itself)
    Therefore, f is surjective

    Thus, f is an automorphism of G.
    QED.

    Now we have to show, by example, that f need not be an automorphism if G is infinite.
    Consider the set of integers Z, under the operation of addition. It is easily shown that Z forms a group under addition.
    Also, Z has no element of order 2:
    Since the operation is addition, "g^2" is actually g + g = 2g.
    The identity element is 0.
    For any g in Z, 2g = 0 implies that g = 0. Thus there is no element of order 2.
    Now consider the above function f. In Z, it is defined as:
    f(g) = 2g, for all integers g
    However, f is not an isomorphism, since it is not surjective.
    For example, 3 is in Z, but there is no element g in Z such that 2g = 3.
    That is, 3 has no preimage under f.
    (Or in other words, f(Z) is not equal to Z. It is, rather, a proper subset of Z, namely the set of even integers)
    Hence, f is not an automorphism on Z.

    Definitions and proofs:
    In the above, I have freely assumed that you know all the definitions and results I used. Below I give some of these in case you are not familiar with them:

    Conventions:
    In the following, 'e' denotes the identity element of the group
    'G' is a group

    Definitions:

    1. Order of an element:
    The order of an element g of G is the least positive integer n, if it exists, such that g^n = e. If no such n exists, then the element g is said to have infinite order.
    Note: This is the order of an element in the group, not the order of the group itself.
    "Least positive integer" - this does not include 0. For all g in G, g^0 is always equal to e.

    Refer: http://en.wikipedia.org/wiki/Order_%28group_theory%29

    2. Homomorphism:
    Let G and H be two groups. Let . be the composition operator of G, and let * be the composition operator of H.
    Then a mapping f: G -> H (read: a mapping f from G to H) is said to be a homomorphism iff:

    For all elements a, b in G, f(a.b) = f(a)*f(b)

    Properties:
    If e' is the identity of H, then f(e) = e'
    Let "^-1" denote the inverse of an element. Then:
    For all g in G, f(g^-1) = (f(g))^-1
    That is, the image of the inverse of an element is the inverse of the image of the element.

    Note: f(a) and f(b) are in H, therefore we must use the operator *, and not .
    It is common to drop the operators . an *, and simply write f(ab) = f(a)f(b), since there is no ambiguity in this.

    Refer: http://en.wikipedia.org/wiki/Homomorphism

    3. Kernel (of a homomorphism):
    The kernel of a homomorphism f : G -> H is the set of all elements of G which map to the identity element of H
    Let e' be the identity element of H. Then the kernel of f is:
    ker f = {g in G | f(g) = e'}

    Note that f(e) = e' (always), so that e is always an element of ker f.
    It can be easily proven that the kernel is a subgroup of G.

    Refer: http://en.wikipedia.org/wiki/Group_homomorphism#Image_and_kernel
    http://en.wikipedia.org/wiki/Kernel_%28algebra%29#Group_homomorphisms

    4. Isomorphism:
    An isomorphism is a homomorphism that is bijective (injective and surjective).
    That is:
    For all a, b in G, f(a) = f(b) implies that a = b (Injective)
    For all h in H, there exists a g in G, such that f(g) = h (Surjective)

    Refer: http://en.wikipedia.org/wiki/Isomorphism

    Proofs:

    1. If G is an Abelian group, then, for all a, b in G, (a^2)(b^2) = (ab)^2
    (a^2)(b^2) = (aa)(bb)
    = a(ab)b [Associativity]
    = ab(ab) [Since G is Abelian, (ab)b = b(ab)]
    = (ab)(ab) [Associativity]
    = (ab)^2

    QED.

    2. A homomorphism is injective if and only if its kernel is trivial:
    Let f: G -> H be a homomorphism, and let e' be the identity of H.
    One part is easy to prove - If the kernel is trivial, then the homomorphism is injective:
    Suppose the kernel is trivial.
    Then, for any g in G, f(g) = e' implies that g = e [Since f(g) = e' implies g is in ker f = {e}]
    Let a, b in G be such that f(a) = f(b)
    f(a(b^-1)) = f(a)f(b^-1) [Since f is a homomorphism]
    f(a(b^-1)) = f(a)((f(b))^-1) [Since f is a homomorphism]
    f(a(b^-1)) = f(a)((f(a))^-1) [Since f(a) = f(b)]
    f(a(b^-1)) = e'
    Therefore, a(b^-1) = e
    Which implies that a = b
    Thus, f(a) = f(b) implies that a =b
    Hence, if the kernel is trivia, then the homomorphism is injective.

    The other part is easier to prove - If a homomorphism is injective, then its kernel is trivial:
    Suppose the homomorphism, f, is injective:
    Then, for all a,b in G, f(a) = f(b) implies that a = b
    We know that f(e) = e'
    Hence, e is in ker f.
    Let g be any element in ker f. Then, f(g) = e'
    But f(e) = e'
    Which implies that f(g) = f(e)
    Which implies that g = e [Since f is injective]
    Therefore, if g is in ker f, then g = e.
    Thus, ker f = {e}.
    Hence, if a homomorphism is injective, then its kernel is trivial.

    Thus, a homomorphism is injective if and only if its kernel is trivial.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 7, 2022, 7:21 pm ad1c9bdddf>
    https://brainmass.com/math/linear-transformation/finite-abelian-group-359983

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