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Prove that in $S_4$ there are $3$ groups that are isomorphic to $D_4$.

I know that the $2$-sylows of $S_4$ should be subgroups of order $8$, but to prove it is a bit tricky for me

Any help would be appreciated, thanks.

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    $\begingroup$ Welcome to the site! It is usually expected that questions provide some context and motivation for the problem. You can edit your question to add such. Absent such information you run the risk of your question being not well-received. $\endgroup$ – quid Jul 11 '15 at 12:57
  • $\begingroup$ What are the $2$-Sylows of $S_4$? $\endgroup$ – Clément Guérin Jul 11 '15 at 13:02
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    $\begingroup$ I know that the 2-sylows of S4 should be subgroups of order 8, but to prove it is a bit tricky for me.. $\endgroup$ – lfc Jul 11 '15 at 13:06
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    $\begingroup$ Try numbering the corners of the square in irregular ways. The symmetry group is (isomorphic to) $D_4$ irrespective of the numbering, but different numberings may result in different subsets of $S_4$. Alternatively you can just conjugate the version of $D_4$ you are familiar with by a 3-cycle. $\endgroup$ – Jyrki Lahtonen Jul 11 '15 at 13:11
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We can see that any Sylow $2$-subgroup is isomorphic to $D_4$ by considering the group of symmetries of the square. By the Sylow theorems the number of such subgroups is congruent to $1$ mod $2$ and divides $24/8=3$, so there is either only one such subgroup or there are $3$. If there were only one, then this subgroup would contain all elements of order $2$ (because every subgroup of order dividing $8$ is contained in a subgroup of order $8$, and the elements of order $2$ generate subgroups of order $2$). However, the elements $(12),(23),(34)$ are of order $2$ and they generate the whole group, so they cannot all be contained in the same subgroup of order $8$. Thus there are $3$ subgroups isomorphic to $D_4$.

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$\require{AMScd}$Hint: View $D_4$ as the symmetries of a square. Draw a square and label the corners. You will, for example, have the element $(1234)$ that rotates the square: $$\begin{CD} 1 @>>> 2\\ @AAA @VVV\\ 4 @<<< 3 \end{CD} $$

So you get the subgroup with the elements $(1), (12)(34), (14)(23), (1234), (1432), (13)$ and two more elements. Can you find them?

After you have found the two elements, try to see if you can find other subgroups that basically have the same elements. You just relabel the cornors of the square. The elements will be of the same type. So you might have a group that has the element $(12)$.

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