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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!

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  • $\begingroup$ If the $p_i$ were all distinct, then by 11.5 (and induction) the whole group would be cyclic. The order of arguing is not good, one should say this before picking out a product of two. $\endgroup$ – André Nicolas Mar 12 '14 at 23:48
  • $\begingroup$ A cyclic group of order $p^k$ has a subgroup of order $p$. We can think of the group operation as addition modulo $p^k$. Then $p^{k-1}$ generates a cyclic subgroup of order $p$. $\endgroup$ – André Nicolas Mar 12 '14 at 23:53
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Probably it is better if you read only the needed theorems of a proof for the first time, and try to figure out the rest by your own.

Let's try the contrapositive: assume that $G$ does not contain a subgroup isomorphic to $\Bbb Z_p\times\Bbb Z_p$. Then, in the form of Thm 11.12, $G\cong\Bbb Z_{p_1^{r_1}}\times\Bbb Z_{p_2^{r_2}}\times\dots,\ $ we must have $p_i\ne p_j$ for all $i\ne j$.

But, in this case, applying Thm 11.5 and induction we get that $G$ is cyclic.

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