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**slevvio** Let $\displaystyle C_n$ denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order $\displaystyle p^2$ is isomorphic to one of the two groups $\displaystyle C_{p^2}, C_p \times C_p $.

OK!

Let G be an abelian group, with $\displaystyle |G|=p^2$. Suppose G is not isomorphic to a cyclic group of order $\displaystyle p^2 $. So we need to show that $\displaystyle G \cong C_p \times C_p$.

Let $\displaystyle g \in G $ with $\displaystyle g \not= 1_G $. So |g| = p. |g| cannot be $\displaystyle p^2 $ because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order $\displaystyle p^2 $.

Let $\displaystyle h \in G \setminus <g>$. Then |h| = p.

Define $\displaystyle \theta : <g>\times <h> \rightarrow G$, given by $\displaystyle \theta (g^i,h^j) = g^i g^j$.

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 $\displaystyle |<g> \times <h> | = |G| = p^2 $ , so it is only required to prove injectivity or surjectivity, but why is $\displaystyle |<g> \times <h> | = |G| = p^2 $ true?

In trying to work out the kernel it seems that it is only equal to {1} if $\displaystyle <g> \cap <h> = \{1\} $ but is this true? Any help would be appreciated here. Thanks very much