Sylow $2$-subgroup of $S_{2^n}$

Question

I would like to study Sylow 2-subgroups of $S_{2^n}$. I studied Sylow 2-subgroups of $D_n$, but I couldn't find reference for $S_{2^n}$. Could anyone explain about Sylow 2-subgroups of $S_{2^n}$?

Thank you.

Answer 1

I will explain the $p$-Sylow subgroup of any $S_n$. This requires understanding wreath powers.

Consider a perfect $k$-level $p$-ary rooted tree. That is, each node has $p$ child nodes, starting with one node at level $0$ and a total of $p^k$ nodes at level $k$. Below we have a $2$-level $3$-ary rooted tree depicted.

Label the nodes at the bottom level $1$ thru $p^k$. Any automorphism of the tree corresponds to a permutation of the labels $\{1,\cdots,p^k\}$. Consider all automorphisms that come from cycling, in order, all leaves emanating from a given node (rather than an arbitrary permutation of them).

In the picture above, we get three separate cyclic groups generated by $(123)$, $(456)$, $(789)$ (which then form an internal direct product within $S_9$), but cycling the nodes directly below the root (and dragging their children with them) yields a cyclic group generated by $(147)(258)(369)$. The direct product of the first three cyclic groups and the latter cyclic group form a semidirect product, $(C_3\times C_3\times C_3)\rtimes C_3$, within $S_9$. This is called a wreath product, denoted $C_3\wr C_3$.

If a group $H$ acts on a set $\Omega$ used to index copies of $G$ in a direct product $\prod_\Omega G$, then the wreath product $G\wr_\Omega H$ is a semidirect product of $\prod_\Omega G$ and $H$, where conjugating $\prod_\Omega G$ by $H$ amounts to $H$ permuting the coordinates of $\prod_\Omega G$ according to how $H$ acts on $\Omega$. Thus, it has cardinality $|G\wr_\Omega H|=|G|^{|\Omega|}|H|$. If the set $\Omega$ is understood from context it can be dropped from the notation to write $G\wr H$.

In general th cycling automorphisms generate a subgroup which is a wreath power

$$ C_p^{\wr k} : = \underbrace{C_p\wr C_p\wr C_p\cdots\wr C_p}_k $$

(wreath products are associative up to isomorphism, so I didn't bother parenthesizing). By induction and the geometric sum formula, this has cardinality $|C_p^{\wr k}|=p^{p^{k-1}+\cdots+p+1}=p^{(p^k-1)/(p-1)}$.

The $p$-Sylow subgroup of $S_n$ can be determined from $n$'s base-$p$ expansion:

$$ n = \sum a_k p^k \\ P = \prod a_k C_p^{\wr k} $$

Above, we adopt the notation $mG:=\overbrace{G\times\cdots G}^m$ for notational suggestiveness. To see how $P$ is a subgroup of $S_n$, consider drawing $a_k$-many perfect $k$-level $p$-ary rooted trees then labelling all of their lowest level nodes $1$ thru $n$. To see it's a Sylow subgroup, check it's the correct size using Legendre's formula.