Math Help - Homomorphism & Isomorphism

1. Homomorphism & Isomorphism

Hi,

i was wondering if anyone could guide me through this question:

a) Suppose G is an abelian group, and let f:G->G be defined by f(x)=x^3
i) show that f is an isomorphism if |G| is not divisible by 3.

b) give an example to show that f:x->x^3 need not be a homomorphism if G is non-abelian.

c) suppose that G is a group in which x^2=e for all x in G. Prove that G is abelian.

Thanks!

2. Originally Posted by choo
Hi,

i was wondering if anyone could guide me through this question:

a) Suppose G is an abelian group, and let f:G->G be defined by f(x)=x^3
i) show that f is an isomorphism if |G| is not divisible by 3.

b) give an example to show that f:x->x^3 need not be a homomorphism if G is non-abelian.

c) suppose that G is a group in which x^2=e for all x in G. Prove that G is abelian.

Thanks!
For (a), look at the kernel of the map.

For (b), look at $S^3$. The set that you will get given is not a group!

(c) is a classic question that must be asked here 2 or 3 times a month. The trick is to notice that if you take two elements $a, b \in G$ then $(ab)^2 = 1$. What does this tell you?

3. [quote=Swlabr;480558]For (a), look at the kernel of the map.

but this implies only injective. What about surjective??

4. [quote=1234567;482325]
Originally Posted by Swlabr
For (a), look at the kernel of the map.

but this implies only injective. What about surjective??
The question implies the group is finite (the result does not holds otherwise). If $G$ is finite and $\phi : G \rightarrow G$ is an injection, then it holds that $\phi$ is a surjection too.

Can you see why this is? Essentially, it is because of the orders. You know that the pre-image and the image have the same order, and every element of the pre-image maps to a unique element of the image. So it really has to be a surjection. (The converse also holds).

5. [quote=Swlabr;482340]
Originally Posted by 1234567

The question implies the group is finite (the result does not holds otherwise). If $G$ is finite and $\phi : G \rightarrow G$ is an injection, then it holds that $\phi$ is a surjection too.

Can you see why this is? Essentially, it is because of the orders. You know that the pre-image and the image have the same order, and every element of the pre-image maps to a unique element of the image. So it really has to be a surjection. (The converse also holds).
Thank for the explaination.