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I am trying to find the number of Sylow 2-sbgroups of the symmetric group S5. As $ \lvert S_5 \rvert =120=2^3 \cdot 3 \cdot 5$. It has 2-SSG, 3-SSG, 5-SSG. But how to calculate there numbers? What is the general formula to calculate it.

vonbrand
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Sajan Phutela
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    Since a Sylow $2$-subgroup of $S_5$ fixes a unique point, it might help to observe that the total number is equal to five times the number of Sylow $2$-subgroups of $S_4$. – Derek Holt Feb 20 '16 at 13:00

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There are $15$ Sylow-$2$-subgroups of $S_5$. For a proof see here. It equals $\frac{|S_5|}{|P|}=\frac{120}{8}=15$. By Sylow's theorem we have $n_2(S_5)\in \{ 1,3,5,15\}$, and it is easy to show that $n_2(S_5)\ge 6$, so the result follows.

Dietrich Burde
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