3

I know, $|S_4 |=24 =2^3\cdot3$.

So here the order of sylow $2$ subgroup is $8$ and by the third sylow theorem we can say the number of sylow $2$ subgroups is $1$ or $3$. Then by finding the sylow $2$ subgroups explicitly we can conclude that its $3$. But here we also know the number of sylow $3$ subgroups is $1$ or $4$.

My question is can we find the exact number of sylow $2$ and $3$ subgroups without calculating the the sylow subgroups explicitly, just by element counting of the subgroups, because sometimes we need only the exact number in of sylow subgroups ?

Shankhadeep
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  • The number of Sylow $p$-subgroups is $|G:N_G(P)|$, where $P$ is some specific Sylow $p$-subgroup. This avoids having to calculate them all explicitly - you just need to find one of them and compute its normalizer. – Derek Holt Dec 07 '20 at 14:02
  • Thank you, I got this very useful because this thing can be used for any group. – Shankhadeep Dec 08 '20 at 05:16

1 Answers1

5

Suppose $S_4$ has a unique Sylow $2$-subgroup say $K$. By Sylow Second Theorem, $K$ must be normal in $S_4$. But $S_4$ does not have any normal subgroup of order $8$ (Refer here). Therefore the number of Sylow $2$-subgroups in $S_4$ must be three.

Wang Kah Lun
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  • Thank you. Here for permutation group we know particular normal subgroup as An where n is not 4, so thank you so much for giving this idea, I have one more question, for other groups where this kind of normal subgroup may present , there can we say any tricky observation like this ? – Shankhadeep Dec 08 '20 at 05:12