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Let $G$ be a finite group and $p$ a prime such that $p^\alpha$ divides $|G|$ and $p^{\alpha+1} \nmid |G|$. I know that Sylow $p$-subgroups of $G$ are conjugate to one another but if we have some $p$-subgroups of $G$ with the same cardinality(different from $p^\alpha$) and each of them are isomorphic to some Sylow $p$-subgroup(the Sylow p subgroup can be the same or different) of some other group(group has to be the same)...can we say that they are conjugate in $G$?

I can't think of a good example, but this problem below lead me to the question above: In $S_4$ with $|S_4|=2^33$we have $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ as $2$-subgroups of $G$ isomorphic to $\mathbb{Z} _4$. A Sylow $2$-subgroup in $\mathbb{Z} _4$ has cardinality $2^2$. So the images of $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ are conjugate in $\mathbb{Z} _4$. Can I say that $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ are Sylow $2$-subgroupsconjugate in $S_4$

This is a bad example but do you get what I'm trying to say?

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  • $\begingroup$ So your question is whether two isomorphic $p$-subgroups of a group necessarily have to be conjugate. I'd bet against this, and suspect that a semidirect product would make a counterexample. $\endgroup$ – darij grinberg Apr 14 '14 at 0:43
  • $\begingroup$ The $p$-subgroups are not necessarily isomorphic to each other but they are isomorphic to Sylow $p$-subgroups of some other group. Not necessarily the same one. $\endgroup$ – abe Apr 14 '14 at 0:54
  • $\begingroup$ Those $p$-Sylow subgroups, however, are conjugate and therefore isomorphic. $\endgroup$ – darij grinberg Apr 14 '14 at 0:58
  • $\begingroup$ So you're saying conjugate subgroups are isomorphic? $\endgroup$ – abe Apr 14 '14 at 1:02
  • $\begingroup$ I believe you now. I didn't know conjugate subgroups were isomorphic. Thanks $\endgroup$ – abe Apr 14 '14 at 1:03
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Here's a negative example. Take two groups of order 2 in the dihedral group of 4 elements $D_{2\cdot2}\cong V_4$: $H=\{e,\rho\}$ and $K=\{e,\tau\}$. These are normal in $D_{2\cdot2}$ (index 2) so aren't conjugate, but are isomorphic to Sylow-2 groups in the dihedral group $D_{2\cdot3}$ (pick two reflection subgroups of order 2). So as darij said a semidirect product will work.

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  • $\begingroup$ Thanks,this counter example is perfect $\endgroup$ – abe Apr 14 '14 at 1:27
  • $\begingroup$ Are isomorphic to Sylow 4(?) ??? What do you mean? $H$ is cyclic and $K$ is not... $\endgroup$ – Nicky Hekster Apr 14 '14 at 5:30
  • $\begingroup$ Yikes thanks Nicky, my answer is wrong (and it should've said Sylow-2 not Sylow-4). Let me fix it. $\endgroup$ – rVitale Apr 14 '14 at 14:53
  • $\begingroup$ should be ok now. $\endgroup$ – rVitale Apr 14 '14 at 15:03
  • $\begingroup$ In case one is interested in conjugacy of $p$-subgroups: Notice that $H$ and $K$ are conjugate in the larger group $A_4$. Which subgroups of a $p$-group $P$ can be conjugate if we allow them to be in larger groups $G$? What if $P$ is required to be a Sylow $p$-subgroup of $G$? Does the answer change if we allow $G$ to be infinite? $\endgroup$ – Jack Schmidt Apr 14 '14 at 15:08

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