TITLE: Maximal subgroups of a certain finite 2-group
QUESTION [10 upvotes]: The following came up in a problem on reconstruction of digraphs.  I determined enough about the answer to satisfy the application completely, but still I am curious to know what the complete solution is.  My group theory is weak, so apologies if this is too simple.
Let $T$ be a full binary tree with depth $k$. Call its levels $L_0,\ldots,L_k$.  Here is the case $k=4$:
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The number of leaves is $n=2^k$.
Let $A$ be the full automorphism group of $T$ and let $f$ be its (faithful) action on the leaves of the tree, i.e. on $L_k$. Obviously $f(A)$ is an iterated wreath product of $\mathbb{Z}_2$ with itself and has order $2^{n-1}$.  It is, indeed, the Sylow 2-subgroup of $S_n$.
The problem is: what are the subgroups of $f(A)$ of index 2?
Here is what I think the answer is.  Let $X$ be a union of levels of $T$, including at least one level other than $L_0$. Let $P$ be the set of all $f(\gamma)$ such that $\gamma\in A$ and the action of $\gamma$ on $X$ is an even permutation.  Then $P$ is a subgroup of the desired index. I'm guessing there are no others...

REPLY [9 votes]: You can easily count the number of maximal subgroups of $W(k)$, the $k$-fold iterated wreath product of $\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$
To proceed, note that $W(k) = W(k-1) \wr \mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \wr \mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1) \times E(k-1),x]$ has order $2^{k-1},$ so that the largest elementary Abelian factor group of the original wreath product does have order $2^{k},$ as claimed.
Hence the group $W(k)$ has $2^{k}-1$ maximal subgroups, since there is a bijection between maximal subgroups of $W(k)$ and maximal subgroups of $W(k)/\Phi(W(k)).$