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