Let

denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order

is isomorphic to one of the two groups

.

OK!

Let G be an abelian group, with

. Suppose G is not isomorphic to a cyclic group of order

. So we need to show that

.

Let

with

. So |g| = p. |g| cannot be

because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order

.

Let

. Then |h| = p.

Define

, given by

.

This is a homomorphism (need to use abelian-ness to prove this)

But how do I show that this is surjective and injective? It says here that

, so it is only required to prove injectivity or surjectivity, but why is

true?

In trying to work out the kernel it seems that it is only equal to {1} if

but is this true? Any help would be appreciated here. Thanks very much