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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?

Jyrki Lahtonen
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    If $G$ has $n$ generators of order $2$, then any quotient of $G$ also has $n$ generators of order dividing $2$. So $P_n/[P_n,P_n]$ is an elementary $2$-group of rank at most $n$. To see the rank is $n$ you can argue by induction; once $P_n$ maps to $(\mathbf{Z}/2)^n$, $P_{n+1}$ maps to $(\mathbf{Z}/2)^n \wr (\mathbf{Z}/2)$ which then maps to (\mathbf{Z}/2)^{n+1}$. – user670344 May 05 '19 at 12:44
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    You can also use induction: if $H$ is the abelianization of $G$, then the abelianization of $G \wr \mathbf{Z}/2$ is the abelianization of $H \wr \mathbf{Z}/2$, so it reduces to finding the abelianization of $(\mathbf{Z}/2)^n \wr (\mathbf{Z}/2)$ which is not too hard to show is $(\mathbf{Z}/2)^{n+1}$. – user670344 May 05 '19 at 12:50
  • @user670344 You make it sound trivial! I have clearly not worked enough on abelianizations and wreath products given that I failed to see that inductive argument:-) Feel free to post it as an answer (if/when you find the time). – Jyrki Lahtonen May 05 '19 at 13:11

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