External Direct Product

**slevvio** · January 26th 2010, 08:56 AM

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

OK!

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

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

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

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

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

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

**tonio** · January 26th 2010, 09:42 AM

Originally Posted by slevvio

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

OK!

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

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

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

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

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

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

If $h^a=g^b$ , for some $0\leq a,b< p$ , then as $(a,p)=1$ there exists $k\in\mathbb{Z}$ s.t. $ak=1\!\!\!\pmod p\Longrightarrow h=h^{ak}=g^{bk}\in <g>$ , contradicting your choice of $h$ , and
this simply means $<g>\cap <h>=1$ .

Tonio

**Drexel28** · January 26th 2010, 12:42 PM

Originally Posted by slevvio

But how do I show that this is surjective and injective? It says here that $|<g> \times <h> | = |G| = p^2$ , so it is only required to prove injectivity or surjectivity, but why is $|<g> \times <h> | = |G| = p^2$ true?

Because $\left|\left\langle g\right\rangle\right|=\left|\left\langle h\right\rangle\right|=p$ . So from basic set theory and combinatorics $\text{card }M=m,\text{card }N=n\implies \text{card }M\times n=mn$ . Also, the reason you only have to prove either surjectivity or injectivity is that a surjective function from a set to an equipotent set is injective and vice versa if the sets are finite.

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