Revisions to Maximal subgroups of a certain finite 2-group

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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 $k=4$:

tree with 4 levels http://cs.anu.edu.au/%7Ebdm/tree.png(source)

The number of leaves is $n=2^k$. Let

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...

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 $k=4$:

tree with 4 levels http://cs.anu.edu.au/%7Ebdm/tree.png

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...

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 $k=4$:

tree with 4 levels (source)

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...