# Sylow subgroups of $S_3$ and $S_4$

Find the three 2-Sylow subgroups of $S_3$ and find a 2-Sylow subgroup and a 3-Sylow subgroup of $S_4.$

I just learned Sylow' theorem at the moment and I don't know how to do these problems. I know a p-Sylow subgroup of a group $G$ is a subgroup of order $p^k$, where $p^k$ divides the order of $G$ but $p^{k+1}$ does not. $|S_3| = 6 = 2\dot\ 3$ and $|S_4| = 24 = 2^3\dot\ 3$. How can I solve these types of problems?

$S_3$ is all the different ways you can rearrange the three letters $(a,b,c)$. For instance, the permutation $(a,b,c) \mapsto (b,c,a)$ is one element of $S_3$. A Sylow-2 subgroup will, as you wrote, just be a subgroup of order 2. 2 is prime, so to find such a subgroup, you just have to come up with some way of swapping the letters that has order two, which means if you do it twice, you're back to the start.
$S_4$ is the different ways you can rearrange the four letters $(a,b,c,d)$. A Sylow-3 subgroup has to be generated by some move that will get back to the sequence $(a,b,c,d)$ after 3 moves. Can you think of some moves that will work in each case? There's actually a bunch of them, so it would be good to try and think of a few different ones.
• The size of a Sylow p-subgroup depends on the size of the group. For $S_3$, you know its size is $6 = 2\cdot 3$. The size of a Sylow 2 subgroup is 2 in this case. The size of a Sylow 3 subgroup is 3. But, for $S_4$, the order is $2^3\cdot 3$. So for $S_4$, as Sylow 2-subgroup has order $2^3 = 8$, and a Sylow 3-subgroup has order $3$. You just factor the order into primes, and the power of each prime you see in the factorization is the size of a Sylow subgroup for that prime. – Zach L. Nov 14 '13 at 7:32