prove for f?

Question

Let f: G->G be homomorphism such that f(x) doesn't equal to I, where G is a group of prime order with identity I and x belong to G. Prove that f is an isomorphism.

Answer 1

It's not right.

f: G -> G so that f(x)=1 for every x in G is a homomorphism but it
isn't a homomorphism.

If f isn't trivial you can prove the statement.
f homomorphism => f(G) and Kerf are subgroups of G.
G of prime order implies (according to Lagrange's theorem) that its only subgroups are {1} and G itself.
If f isn't trivial Kerf /= G (not equal) so Kerf={1} and f is injective.
On the other side f not trivial implies f(G) /= {1} therefore we have
f(G)=G and f is surjective. So f is a isomorphism.

Excuse me, but my English sucks :-)

Answer 2

I'd have to work to hard to review the material.

Maybe this will help:
http://en.wikipedia.org/wiki/Homomorphism