Show that the map $f(a)=a^2$ is an isomorphism of groups from G to itself.

Question

Let $G$ be a commutative group of odd order.

$f(ab)=(ab)^2=a^2b^2=f(a)f(b)$ so clearly its a homomorphism.

Now suppose $f(a)=f(b)$.

Then $a^2=b^2$.

Multiply on the left by $a$.

$a^3=ab^2$ and since G is odd, $a^3 =e$.

So $e=ab^2$. Then multiplying by $b$ on the right,

$eb=ab^3$ and $b^3 = 1$ so $a=b$. It is injective.

How do we show it is surjective?

You are not given $a^3=e$, so your proof of injectivity is flawed. Anyway, once you proved injectivity it is just counting elements. — user10354138, Jun 12 '19 at 15:57
To show injectivity of homomorphisms is somewhat easier than for general functions. You may actually assume that $b=e$ and just show $f(a)=e\implies a=e$. — Arthur, Jun 12 '19 at 15:59
Actually, every element does have odd order because its order divides the order of the group which is assumed to be odd. — Somos, Jun 12 '19 at 16:23

score 4 · Accepted Answer · answered Jun 12 '19 at 16:02

4

What's the kernel? If $a^2=e$, then $a=e$. Otherwise $\vert a\vert=2 \Rightarrow \Leftarrow $.

answered Jun 12 '19 at 16:02

How do we get injectivity from this? – s_healy Jun 12 '19 at 16:12
1

See @Arthur's comment. – Jun 12 '19 at 16:16

Chinnapparaj R · Answer 2 · 2019-06-12T16:05:56.270

2

Your answer is not correct $$a^2=b^2 \implies a^2b^{-2}=e \implies (ab^{-1})^2=e$$ so order of $ab^{-1}$ is either one or two and by hypothesis, it is one and so $a=b$

Now the map from a finite set to itself is one to one, so it is surjective as well!

edited Jun 12 '19 at 16:05

answered Jun 12 '19 at 16:00

Chinnapparaj R

11,989

Show that the map $f(a)=a^2$ is an isomorphism of groups from G to itself.

2 Answers2