Let $N\rtimes Q$ be a semidirect product of groups with normal subgroup $N$ and quotient $Q$.

Let $H\le N$ be a fixed subgroup of finite index $d$. Is it possible to classify the subgroups $U$ of $N\rtimes Q$ with index $d$ such that $U\cap N = H$?

By classify, I would like to at least know that if such a $U$ exists, must it be unique? and if it is not unique, then is the set of all $U$'s satisfying the condition a torsor under some (cohomology?) group?

If this is difficult in general, can we at least do it if we assume $d = 2$?

  • $\begingroup$ No it is not necessarily unique. Consider for example a dihedral group $C_n\rtimes C_2$. If $H$ is a subgroup of $C_n$, then any reflection $t$ will yield a subgroup $H\rtimes \langle t\rangle$ of the appropriate index, and there will be more than one such subgroup if $H$ is proper. $\endgroup$ – verret Sep 11 '18 at 5:18
  • $\begingroup$ In fact, this works whenever $N$ is cyclic, because in that case $H$ is characteristic in $N$ so normalised by any complement $Q$. So, as you run over the different complements for $N$, you will get many possibilities for $U$. $\endgroup$ – verret Sep 11 '18 at 5:20

Let's consider the special case when $H \unlhd N$ - the general case is more complicated. Then subgroups $U$ of the required type exist if and only $H \unlhd G$, and in that case one such subgroup is $H \rtimes Q$. The set of all such subgroups $U$ corresponds in a fairly obvious way to the set of complements of $N/H$ in $G/H$. Note that not all of these subgroups are necessarily split extensions.

In particular, if $N/H$ is abelian, then the conjugacy classes of subgroups $U$ with this property is in one-one correspondence with the first cohomology group $H^1(Q,N/H)$, which can typically be computed. Indeed, this technique is used as a component of computer algorithms to find all conjugacy classes of subgroups of a given finite group.

  • $\begingroup$ I think you swapped $H$ and $N$? I assume $G := N\rtimes Q$? $\endgroup$ – stupid_question_bot Sep 11 '18 at 15:56
  • $\begingroup$ I have reswapped $H$ and $N$. $\endgroup$ – Derek Holt Sep 11 '18 at 17:51

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