Maximal subgroups of a certain finite 2-group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:02:15Z http://mathoverflow.net/feeds/question/106164 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106164/maximal-subgroups-of-a-certain-finite-2-group Maximal subgroups of a certain finite 2-group Brendan McKay 2012-09-02T08:15:58Z 2012-09-08T11:38:27Z <p>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.</p> <p>Let $T$ be a full binary tree with depth $k$. Call its levels $L_0,\ldots,L_k$. Here is the case $k=4$:</p> <p><img src="http://cs.anu.edu.au/~bdm/tree.png" alt="tree with 4 levels"></p> <p>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 $S_n$.</p> <p>The problem is: what are the subgroups of $f(A)$ of index 2?</p> <p>Here is what I <em>think</em> 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...</p> http://mathoverflow.net/questions/106164/maximal-subgroups-of-a-certain-finite-2-group/106167#106167 Answer by Geoff Robinson for Maximal subgroups of a certain finite 2-group Geoff Robinson 2012-09-02T08:50:59Z 2012-09-02T17:12:33Z <p>You can easily count the number of maximal subgroups of $W(k)$, the $k$-fold interated wreath product of <code>$\mathbb{Z}_{2},$</code> 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 <code>$W(k) = W(k-1) \wr \mathbb{Z}_{2}.$</code> Factor out the Frattini subgroup of the base group, and by induction, you are left with <code>$E(k-1) \wr \mathbb{Z}_{2}$</code>, 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 $W(k)/\Phi(W(k)).$</p>