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Question

let G be a group of prime order with identity e. Suppose f:G --> G
is a homomorphism such that f( a ) not equal to e for some, a, that belongs to G. completely prove that f is an isomorphism.

Answer 1

yes and no

Answer 2

Let say f trivial if f: G -> G so that f(x)=1 for every x belongs to G.

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 an isomorphism.

Excuse me, but my english is so weak.