# Group homomorphism

1. Feb 28, 2008

### strangequark

1. The problem statement, all variables and given/known data
Suppose that $$\phi$$ is a homomorphism from a finite group G onto G' and that G' has an element (g') of order n. Prove that G has an element of order n.

2. Relevant equations
for a homomorphism,
1) $$\phi(a*b)=\phi(a)*\phi(b)$$
2) $$\phi(a^{n})=(\phi(a))^{n}$$
3) $$\phi(e_{G})=e_{G'}$$

3. The attempt at a solution

It is clear to me that G will contain some non-identity element, say g, which is the preimage of g'. By property 2) that I listed above, $$g^{8}$$ is obviously an element of the kernal of G, and the homomorphism is not the trivial map because $$g^{n}$$ for 0<n<8 is not the identity in G and doesn't map to the identity in G'. Basically, I'm seeing that $$g^{8}$$ maps to the identity in G', but I don't understand why this implies that $$g^{8}=e$$...

I would really appreciate a kick in the right direction... thanks

2. Feb 28, 2008

### Mystic998

You're completely right. It doesn't imply g^8 = e. You're going to have to do a bit more work than just finding an element of the preimage of g'. Unfortunately, I'm having trouble coming up with the exact proof off the top of my head, so I can't give you much of a kick in the right direction. Look at some properties of preimages of subgroups maybe?

3. Feb 28, 2008

### rrogers

Let T(a)^8=I', then a has to have an order n*8, and (a^n)^8=I
This is also summarized by some subgroup divisibility theorem; but it's been to long for me to remember.
For an Image: Imagine a large gear driving a small gear but every tooth of the large gear strikes the same tooth of the small gear on every revolution.