# finite Abelian group

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).

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#### Solution Preview

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 ...

#### Solution Summary

The solution shows 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).