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See page 4 for the proof on the existence of a Sylow $p$-subgroup

I followed the algorithm presented in the proof to build a subgroup with $8$ elements: $$(1), \, (1 2), \, (3 4), \, (1 2)(3 4), \, (1 3)(2 4), \, (1 4)(2 3), \, (1 4 2 3), \, (1 3 2 4)$$ To summarize, I began with $H_1 = \{ \, (1), \, (1 2) \, \}$ and from the normalizers of $H_1$, pick out two additional elements to make $H_2$, which has four elements. Then from the normalizers of $H_2$, pick out four more elements to complete the subset ....

It was good exercise to see how the proof works, but it took so darn long. Now, I still need to conjugate my initial subgroup to find the other subgroups. Is there a better way to do this?

Andy Tam
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For $S_4$ you may notice that the action of the dihedral group $D_8$ on the 4 vertices of the square gives an embedding $D_8\to S_4$, which must be a Sylow 2-subgroup. This subgroup is not normal, so using Sylow theory you conclude there are 3 conjugates. These are not hard to write down.

David Hill
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