# External Direct Product

• Jan 26th 2010, 08:56 AM
slevvio
External Direct Product
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
• Jan 26th 2010, 09:42 AM
tonio
Quote:

Originally Posted by 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

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

Tonio
• Jan 26th 2010, 12:42 PM
Drexel28
Quote:

Originally Posted by slevvio

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?

Because $\displaystyle \left|\left\langle g\right\rangle\right|=\left|\left\langle h\right\rangle\right|=p$. So from basic set theory and combinatorics $\displaystyle \text{card }M=m,\text{card }N=n\implies \text{card }M\times n=mn$. 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.