# Thread: Sylow 3-subgroup of S_4

1. ## [Solved] Sylow 3-subgroup of S_4

The number of m-cycle in $S_{n}$ (n>=m) is given by
$\frac{n(n-1)(n-2)..(n-m+1)}{m}$.

For $S_{4}$, the groups generated by all 3-cylces (order 3) that I found are

<(1 2 3)>, <(1 2 4)>, <(1 3 4)>, <(2 3 4)>,
<(1 3 2)>, <(1 4 2)>, <(1 4 3)>, <(2 4 3) >

According to the third Sylow theorem, the number of Sylow 3-subgroups is congruent 1 modulo p and divides |G|. Since |G|=24 and p=3, the number of Sylow 3-subgroup should be only 4.

I think the above eight groups generated by 3-cycles are all Sylow 3-subgroups and the number is 8. What went wrong in my procedure?

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My mistake. Four subgroups are right. <(1 2 3)>, <(1 2 4)>, <(1 3 4)>, <(2 3 4)>
Other groups are equivalent to those four subgroups.