I started thinking about this for no particular reason.
Let $P_n$ be a Sylow $2$-subgroup of the symmetric group $S_{2^n}$. What does its abelianization $P_n/[P_n,P_n]$ look like?
The groups $P_n$ have the well known recursive structure as a wreath product tower: $P_n\simeq P_{n-1}\wr C_2$. $P_n$ is also isomorphic to the group of graph automorphisms of a full rooted binary tree of depth $n$ (so $2^n$ leaves). See for example here.
My thinking: The group $P_n$ is generated by the following permutations $g_i,i=1,2,\ldots,n$, $$ \begin{aligned} g_1&=(12),\\ g_2&=(13)(24),\\ g_3&=(15)(26)(37)(48),\\ \vdots&\\ g_n&=\prod_{j=1}^{2^{n-1}}(j;j+2^{n-1}), \end{aligned} $$ where $g_j$ is a product of $2^{j-1}$ disjoint $2$-cycles pointwise intechanging the ranges $[1,2^{j-1}]$ and $[2^{j-1}+1,2^j]$.
All the generators $g_j$ have order two, so it seems to me that $P_n/[P_n,P_n]$ should be an elementary abelian $2$-group. I have a hunch that the rank of this elementary abelian $2$-group should be equal to $n$.
- Basically this is because I don't see how there could be relations stopping one of the generators from having an impact on the abelianization, and
- in the cases $n=1$ and $n_2$ with $P_1\simeq C_2$ and $P_2\simeq D_4$ (=the dihedral group of symmetries of a square) it is easy to verify by hand that $$P_n/[P_n,P_n]\simeq C_2^n.$$
Is my hunch correct? Why or why not?
