I've been stuck on this problem for a while now.

Let $G$ be finite and abelian. Suppose $\exists x \in G$ such that $x$ has non-square-free order, i.e., $|x| = p^km$ with $p$ prime and gcd$(p,m)=1$ and $k > 1$. Then, show that there exists an element of order $p^2$.

My attempt (using the erroneous assumption that $|G|$ was square-free, instead of $|x|$):

Let $G$ be as above. By Sylow, there exists a subgroup $H$ with order $p^2$ as $p^2$ divides the order of $G$

Furthermore, as $|H| = p^2$, $H$ is abelian. Thus, $H \cong C_p \times C_p$ or $H \cong C_{p^2}$.

I'm not sure how to proceed here; I had the thought that, if I was able to show that H was cyclic, I'd be done. We have the fundamental theorem of abelian groups and Sylow, but not too much else.

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    $\begingroup$ I don't think this is true. $\mathbb Z/p\mathbb Z\times\mathbb Z/p\mathbb Z$ has order $p^2$ and any (nonidentity) element $(n,m)$ has order $p$ since $n$ has order $p$ or $m$ has order $p$ and the order of $(n,m)$ is the lcm of $1,p$ or $p,p$. $\endgroup$ – Ian Coley Feb 14 '14 at 8:45
  • $\begingroup$ @IanColey I apologize, I misread the question. The assumption is that an element has non-square-free order. $\endgroup$ – Lost Feb 14 '14 at 8:46
  • $\begingroup$ Could you edit your question to reflect that? $\endgroup$ – Ian Coley Feb 14 '14 at 8:47
  • $\begingroup$ @IanColey Done, thanks for the catch - I'll have to see where this leads me, but any suggestions would be helpful $\endgroup$ – Lost Feb 14 '14 at 8:49

Take the element $y=x^{p^{k-2}m}$. Then $y^{p^2}=e$. Further, suppose $y^n=e$. Then $$ e=y^n=x^{p^{k-2}mn}\implies p^km\mid p^{k-2}mn\implies p^2\mid n $$ since may choose $k$ so that $(p,m)=1$.

  • $\begingroup$ Sorry this was less hinty than it might've been. This question basically boils down to the fundamental theorem of cyclic groups applied to $\langle x \rangle$. $\endgroup$ – Ian Coley Feb 14 '14 at 8:52
  • $\begingroup$ No, it's fine, thanks - I have several other parts to prove, so still got my work cut out for me. I follow what you're saying, except that I'm not quite sure what you mean by that last line. How does $k$ have anything to do with gcd$(p,m)$? The assumption (which I forgot to write, sorry) is that $p$ and $m$ already are relatively prime $\endgroup$ – Lost Feb 14 '14 at 8:55
  • $\begingroup$ You want to choose $k$ large enough so that $|x|=p^km$ and $(p,m)=1$, i.e. $p^k$ is the largest power of $p$ dividing $|x|$. Without this assumption, we wouldn't be able to just divide $p^k$ into the $p^{k-2}$ part above. $\endgroup$ – Ian Coley Feb 14 '14 at 8:57
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    $\begingroup$ Thanks again for the answer, I was focusing too much on Sylow and forgot about being able to do this. $\endgroup$ – Lost Feb 14 '14 at 9:03

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