Given $G$, $|G|=p^n$, $H,K\le G$ s.t. $|H|=|K|=p^k$ for some $k<n$.

Can we say that $H\simeq K$? I think it's true and I tried to prove it building by hand the isomorphism $\psi:H\longrightarrow K$ and working by induction on $k$: in fact every $p$-group (such $K$ and $H$ are) contains an element of order $p$; call them $h\in H$ and $k\in K$: I can make correspond these two elements in the isomorphism $\psi$.

Then I can consider the quotient groups $H/\langle h\rangle$ and $K/\langle k\rangle$ that are isomorphic by inductive hypotesis... and then? Can I conclude that $H$ and $K$ are isomorphic too?

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    $\begingroup$ No, this fails for plenty of examples (try a few of order $8$). $\endgroup$ – Tobias Kildetoft May 13 '14 at 17:30
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    $\begingroup$ No, consider $G=\mathbb{Z}_4\times \mathbb{Z}_2\times \mathbb{Z}_2$ of order $8$. $\endgroup$ – Dietrich Burde May 13 '14 at 17:31
  • $\begingroup$ I think it might actually be possible to completely classify $p$-groups satisfying this. A first step is noting that it is easy to see when it holds for abelian groups, and for $p=2$ the only other possibility is being generalized quaternion (so one needs to check if those satisfy it). For $p\geq 3$ a non-abelian example will have all elements of order $p$. $\endgroup$ – Tobias Kildetoft May 13 '14 at 17:35
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    $\begingroup$ Take the direct product of two non-isomorphic $p$-groups of the same order. $\endgroup$ – Eric M. Schmidt May 13 '14 at 17:43
  • $\begingroup$ @DietrichBurde:why $\mathbb Z_4\times\mathbb Z_2\times\mathbb Z_2$ has order $8$? Doesn't it have order $16$? $\endgroup$ – Joe May 13 '14 at 17:45

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