# $p$-subgroups conjugate iff $\cong$ to Sylow p-subgroups of some other groups?

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?

• 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. – darij grinberg Apr 14 '14 at 0:43
• 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. – abe Apr 14 '14 at 0:54
• Those $p$-Sylow subgroups, however, are conjugate and therefore isomorphic. – darij grinberg Apr 14 '14 at 0:58
• So you're saying conjugate subgroups are isomorphic? – abe Apr 14 '14 at 1:02
• I believe you now. I didn't know conjugate subgroups were isomorphic. Thanks – abe Apr 14 '14 at 1:03

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.
• Are isomorphic to Sylow 4(?) ??? What do you mean? $H$ is cyclic and $K$ is not... – Nicky Hekster Apr 14 '14 at 5:30
• 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? – Jack Schmidt Apr 14 '14 at 15:08