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Let $G$ be a finite non-cyclic $p$-group of order $p^n$, $n >1$, where $p$ is odd prime. I need to prove(By elementary methods)that $G$ has a subgroup isomorphic to $\Bbb{Z}_p \times \Bbb{Z}_p$?

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  • $\begingroup$ This is not true unless you also assume that $p > 2$. $\endgroup$ – Tobias Kildetoft Sep 30 '16 at 10:29
  • $\begingroup$ why? i think it is correct $\endgroup$ – Rima Sep 30 '16 at 10:30
  • $\begingroup$ Then you should look at some examples of $2$-groups. $\endgroup$ – Tobias Kildetoft Sep 30 '16 at 10:31
  • $\begingroup$ @TobiasKildetoft Also I think it is correct for any prime $\;p\;$. Do you have some counter example in mind? $\endgroup$ – DonAntonio Sep 30 '16 at 10:31
  • $\begingroup$ @DonAntonio All the generalized quaternion groups (these are in fact precisely the counter examples). $\endgroup$ – Tobias Kildetoft Sep 30 '16 at 10:32
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If $G$ is abelian, then $G$ has at least two distinct subgroups of order $p$ such as $H,K$. Thus $H \times K \cong \Bbb{Z}_p \times \Bbb{Z}_p $.

Now let $G$ is non-abelian. We use induction on $n$. If all maximal subgroups of $G$ are cyclic, then $G \cong Q_8$, a contradiction. Thus there exists a non-cyclic maximal subgroup of $G$ such as $H$. By induction hypothesis $H$ has a non-cyclic subgroup of order $p^2$. Thus $G$ has a non-cyclic subgroup of order $p^2$.

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  • $\begingroup$ Under what assumption on an abelian group $G$ are you assuming it has at least two distinct subgroups of order $p$? $\endgroup$ – Nex Oct 1 '16 at 18:35
  • $\begingroup$ @Nex Because if $G$ is a finite abelian $p$-group such that $G$ has exactly one subgroup of order $p$, then $G$ is cyclic. $\endgroup$ – Rima Oct 2 '16 at 3:24

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