I am trying to find all the Sylow 2 subgroups of S4 using Sylow’s theorems. Now, I know that a Sylow 2 subgroup of S4 has size 8, and that there are either 1 or 3 of them (as the number of of Sylow 2-subgroups has form 1+2k and divides 3, the index).

Now my lecturer states “stabilisers of the 3 different bisections of {1,2,3,4} yield 3 distinct Sylow 2 subgroups”

Now my questions are:

1) What does it mean by stabilisers of 3 different bijections of {1,2,3,4}? Is that to say the elements of S4 which send {1,2} and {3,4} to {1,2},{3,4} i.e. by (12)(34)?

2) How does he know that there are 8 elements of this set? Perhaps this will become clearer after the 1st question is answered.

Many thanks, group theory is hard.

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    $\begingroup$ @MarkBennet i goofed, i meant bisections, thanks $\endgroup$ – jacobe Jan 20 '19 at 17:08

1) Yes, you've got it. There are three unique ways to write $\{1,2,3,4\}$ as a union of two sets of size two. These are $\{1,2\}\cup\{3,4\}$, $\{1,3\}\cup\{2,4\}$, and $\{1,4\}\cup\{2,3\}$. These should not be called "bijections of $\{1,2,3,4\}$", but this is what was probably meant. Let's look at the first one, the one you mentioned: $\{1,2\}\cup\{3,4\}$. We can consider the subgroup $H$ of $S_4$ consisting of permutations that preserve this decomposition, i.e., which send $\{1,2\}$ to either $\{1,2\}$ or $\{3,4\}$. You can see $$H = \{(1), (1 2), (3 4), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 3 2 4), (1 4 2 3)\}.$$ 2) It's clear there are going to be 8 elements in $H$, since we have 4 choices for where $1$ is sent, then we have no choice for where to send $2$ (it must stick next to where $1$ is sent), and then we have $2$ choices for where to send $3$, and then no choice for $4$.

So $H$ is a 2-Sylow subgroup of $S_4$. Two others come from the other two decompositions. There can't be more than three of them, as you said. So these are all of them.

In case you're curious, you can show that $H$ (and thus all of the 2-Sylow subgroups) is isomorphic to $D_4$, the dihedral group of order 8. You can see this pretty concretely if you know how $D_4$ acts on the vertices of a square by rotation and reflection. If you label the vertices of a square clockwise by $1$, $3$, $2$, and $4$, then all of these rotations and reflections keep opposite corners together, i.e., they do what elements of $H$ did. The other 2-Sylow subgroups come from labeling the vertices differently (and that's exactly the fact that the 2-Sylow subgroups are conjugate in $S_4$, since conjugacy is pretty explicitly just relabeling in the symmetric groups).

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    $\begingroup$ Thanks for clearing this up, I completely get it now :) $\endgroup$ – jacobe Jan 20 '19 at 17:09
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    $\begingroup$ The comment about the geometric interpretation of the situation in terms of the symmetries of a square makes this a particularly nice answer. $\endgroup$ – Travis Willse Jan 20 '19 at 17:27
  • $\begingroup$ "These should not be called 'bijections of $\{1,2,3,4\}$'": You are correct, and it appears Jacobe misread the problem, because it says bisections, which makes more sense. $\endgroup$ – Bob Jones Jan 20 '19 at 22:37

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