I am trying to find out the number of Sylow $3$ subgroups of $S_4$, Order of $S_4$ is $24$,we write $24$ as $24=2*2*2*3$, so we both have sylow $2$ subgroups and sylow $3$ subgroups. Let $n_2$ and $n_3$ are their respective numbers. By using Sylow's 1st Theorem The possible values of $n_3$ are $1,4$ and for $n_2=1,3$.We know $S_4$ has order $3$ for the elements $ (1,2,3),(2,3,4),(3,4,1),(1,2,4)$,Each of them generates distinct cyclic subgroups of $S_4$,Thus we get $n_3=4$.Now we can think each of Sylow $2$ subgroups as the rotation,reflection of a square having vertices $1,2,3,4$ about the axes through the center which is $D_4$.We get such three Sylow $3$ subgroups ie.$n_2=3$.I have solved my problem using the structure of $S_4$,But suppose I have no such idea on the algebraic structure of $S_4$ and trying to do it using only Sylow's Theorems.can I be successful?Is it necessary to study the algebraic structure of any Symmetric group to find it's Sylow $p$ subgroups?

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    $\begingroup$ There are many (nonisomorphic) groups of order 24, and they don't all have the same number of Sylow-3 subgroups – you need to know something about the structure of $S_4$. $\endgroup$ – Gerry Myerson Nov 12 '17 at 10:14
  • $\begingroup$ @Gerrry Myerson Is it sufficient to know only the elements of $S_4$ and their orders to find out number off Sylow $2$subgroups and Sylow $3$ subgroups? $\endgroup$ – Subhajit Saha Nov 12 '17 at 10:20
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    $\begingroup$ Since the 3-Sylows have order 3, the number of 3-Sylows is half the number of elements of order 3. But the 2-Sylows have order 8, and there are several groups of order 8, with different numbers of elements of orders 2, 4, and 8, so that's going to be harder to get, if all you know is the number of elements of each order. $\endgroup$ – Gerry Myerson Nov 12 '17 at 10:33

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