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I'm having a hard time conceptualising Sylow p-subgroups and i'm looking for help with the following problem. I'm more so looking for understanding the Sylow p-subgroups than the other steps of the proof.

Problem. Let $S_4$ be the symmetric group on {1, 2, 3, 4}. Show that the quaternion group of order 8 is not a subgroup of $S_4$.

$S_4$ is of order $4!=24$ and we are looking for a subgroup of order 8. Using sylow p-subgroups we get that $24=2^3\cdot3.$ We see that there are 3 sylow 2-subgroups of order 8. We construct one of them $<(1234),(13)(24)>$ which we see is the dihedral group and it's of order 8 because $(1234)^4=(1234)$ and $(13)(24)^2=2$ so $4\cdot2=8$. Now since every sylow 2-subgroup should be conjugate to eachother we take $sds^{-1}=(13)(24)(1234)(42)(31)=(1432)$ and since $(1432)\in{D_4}$ and so every sylow 2-subgroup of order 8 in $S_4$ should be in the dihedral of order 8. And since the quaternion have six elements with the same square and $S_4$ have less than that there is no subgroup of $S_4$ that is isomorphic to the quaternion group of order 8. Is this correct?

Would another way be to show that the conjugate is not in the quaternion group of order 8? So that for $s\in S_4$ and $q\in Q8$ we have $sqs=x$ where $x\notin Q8?$

Shaun
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Marcus
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    Yes. Alternatively, $S_4$ has only 6 4-cycles, so if $Q_8<S_4$ those must be the elements corresponding to $i,j,k,-i,-j,-k$. But e.g. the product of two 4-cycles is even and hence we cannot have satisfy $ij=k$ etc. (recall $n$-cycles are even iff $n$ is odd). – user10354138 Sep 23 '23 at 14:23
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    You could type that up as an answer, @user10354138. – Shaun Sep 23 '23 at 15:37
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    See also this post, where the argument is (perhaps) shorter. A different proof is given here. – Dietrich Burde Sep 23 '23 at 18:21

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