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Let $|G|=p^m$ for $m \ge 2$. If every subgroup of $G$ of order $p^2$ is cyclic, then $G$ has only one subgroup of order $p$.

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  • $\begingroup$ I would say: Take two different subgroups of order $p$, and use those two to construct a non-cyclic subgroup of order $p^2$. $\endgroup$ – Arthur Jun 2 '15 at 5:38
  • $\begingroup$ What you would need would be that $H_1$ and $H_2$ commute or at least that $hH_ih^{-1} = H_i$ for all $h\in H_{3-i}$, not necessarily that $H_i$ is normal in all of $G$. This is not true in general, but if I were to venture a guess based on just general experience, it would be that in a group of order $p^k$, it is true. (This is not experience in algebra, it's experience in general problem solving. You would need to use that $|G| = p^k$ at some point.) $\endgroup$ – Arthur Jun 2 '15 at 5:45
  • $\begingroup$ Yes! it is enough that $H_1$ and $H_2$ commute. but I'm not sure that this is true. $\endgroup$ – user403172 Jun 2 '15 at 5:55
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It is easy to show that $p$-groups have non-trivial center (this is why you really need your group to be a group of order $p^k$ for some $k$). Hence you have $Z(G)\neq \{1_G\}$ in your situation. Take now an element $z\in Z(G)$ of order $p$. If $x\in G$ is another element of order $p$, denote :

$$H_1:=<z>\text{ and } H_2:=<x> $$

Then I claim that $H_1H_2$ is a subgroup of $G$ which is isomorphic to $H_1\times H_2=\frac{\mathbb{Z}}{p\mathbb{Z}}\times \frac{\mathbb{Z}}{p\mathbb{Z}}$ if $H_1\cap H_2=\{1_G\}$ (essentially $H_1$ normalizes $H_2$ because $z$ commutes with $x$ and $H_2$ normalizes $H_1$ because $z$ commutes with $x$). Because of your hypothesis it cannot be done (we have created a non-cyclic group of order $p^2$). Hence for any $x$ of order $p$ in $G$ you have that $H_1\cap H_2\neq \{1_G\}$ because of the cardinality you must then have $H_1=H_2$, whence the result.

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