Here's what I want to show:
Let $G$ be a finite abelian group. If $G$ is non-cyclic, then $G$ contains a subgroup isomorphic to $Z_p \times Z_p$ for some prime p.
The following attempt is what I gleaned from office hours. Here's what I have:
Proof:
Thm 11.5 says $Z_m \times Z_n$ is cyclic iff m and n are relatively prime
Thm 11.12 says that every finitely generated abelian group G is isomorphic to a direct product of cyclic groups of the form:
$Z_{(p_1^{r_1})} \times Z_{(p_2^{r_2})} \times ...$
where $p_i$ are primes (not nec. distinct) and $r_i$ are positive integers
So, G must have a subgroup isomorphic to a direct product of cyclic groups of the form $Z_(p_1^{r1}) \times Z_(p_2^{r_2})$
Why does this follow from 11.12?
Then by 11.5, we know $p_1 = p_2$. If $p_i$ in 11.12 were all different, then they'd all be relatively prime to each other --> G is cyclic.
Why does this mean that G's subgroup isomorphic to $Z_{(p_1^{r_1})} \times Z_{(p_1^{r_2})}$ has p1 = p2? Seems like this is a general statement about the form of G in Thm 11.12, not the subgroup we're talking about...
All that remains is to find a subgroup of $Z_{(p^{r_1})} \times Z_{(p^{r_2})}$ that's isomorphic to $Z_p \times Z_p$. $\langle p^{(r_1-1)}\rangle \times \langle p^{(r_2-1)}\rangle$ is isomorphic to $Z_p \times Z_p$.
Why is this true? I would think $Z_p \times Z_p$ has $p^2$ elements and $\langle p^{(r_1-1)}\rangle\times \langle p^{(r_2-1)}\rangle$ has $(r_1-1) * (r_2-1)$ elements. Where am I going wrong?
Thanks for the help!