Find all the p-Sylow subgroups of $S_4$

Question

Im having a slight difficulty determining $p$-Sylow subgroups.

I am asked to find all $p$-Sylow subgroups of $S_4$.

Work:

So |$S_4|=4!=24=2 \times 2 \times 2 \times 3$ Thus, I will have $2$-sylow subgroups and $3$-sylow subgroups.

I also know that the order of a $p$-sylow subgroup is the highest power of $p$ that divides the order of the group. So in this case, |$2$-sylow subgroup's|${}= 2^3=8$ (Since $8\mid 24$) and |$3$-sylow subgroup's|${}=3^1=3$ (since 3|24)

I also know I will have $\frac{24}{2^3}=3$, $2$-sylow subgroups and $\frac {24}{3}=8$, 3-sylow subgroups (is this correct so far?)

I am having difficulty actually finding out what these groups are..

Also, I am asked to find each $p$-sylow subgroups Normalizer. I know that the normalizer is $N_{S_4}=\{x \in S_4 \mid xS=Sx\}$ where $S$ is a $p$-sylow subgroup. Is there any quick way of determining the normalizer, or is it just trial and error? Do I need to compute all the combinations permutations to see which elements are in the normalizer of a $p$-sylow subgroup?

Thanks!

Answer 1

You are correct about the size of the Sylow subgroups, but not about their count. The numpre of $p$-Sylow subgroup need not be equal to its index, it may only be a divisor of it. So from this there are 1 or 3 $2$-Sylow subgroups and 1 or 2 or 4 or 8 $3$-Sylow subgroups. We have one additional piece of information: The number is $\equiv 1\pmod p$. This does not add anything for $2$-Sylow subgroups, but for $3$-Sylow subgroups we know that their number is either 1 or 4.

What can the $2$-Sylows look like? All elements of order $3$ in the group $S_4$ are of cycle type $(a\,b\,c)$ and there are 8 of them. As each of these is contained in (at least) one $3$-Sylow group, weconclude that one Sylow group is not enough. So the number of them must be $4$. Also, they consist precisely of the three-element group generated by an element of order $3$ (i.e., $\{\text{id},(a\,b\,c),(a\,c\,b)\}$ for suitable $a,b,c$).

The group $S_4$ has 6 elements of cycle type $(a\,b\,c\,d)$ and of order $4$, as well as 6 of type $(a\,b)$ and 6 of type $(a\,b)(c\,d)$ of order $2$. Again, this is too much for a single subgroup of order $8$, hence there must be three $2$-Sylow subgroups. Each $(a\,b\,c\,d)$ must be contained in (at least) one of them, and with it also $(a\,c)(b\,d)$ and $(a\,d\,c\,b)$. One might go through the trouble to test all possible combinations with other elements of order $2$ or $4$ and check if one obtains a group of order $8$. However, it is easier to just "see" a subgroup of order $8$ in $S_4$: Naturally, $S_4$ acts by permuting a set of size $4$, for example the set of vertices of a square (or put differently: We label the vertices with numbers $1,2,3,4$). The square as a symmetry group of its own, consisting of rotations and reflections. This "happens" to be of order $8$ and hence, being a subgroup of $S_4$, must be a $2$-Sylow group. We obtain the different conjugates by changing the labelling o fthe vertices.