Explore BrainMass

# finite Abelian group

Not what you're looking for? Search our solutions OR ask your own Custom question.

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

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