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A generalized dihedral group, $D(H) := H \rtimes C_2$, is the semi-direct product of an abelian group $H$ with a cyclic group of order $2$, where $C_2$ acts on $H$ by inverting elements.

I know that the total number of subgroups of $D(H)$ is the number of subgroups of $H$ plus the sum of subgroup indices of $H$ ($\sum_{L \leq H}[H : L]$).

But I'm not interested in the actual subgroups, I only need the structures of subgroups of $D(H)$ up to isomorphism. Naturally all the subgroups of $H$ are (normal) subgroups of $D(H)$ and for each $L \leq H$ also $D(L) \leq D(H)$. But is that all or can there be other structures as well?

I read about the subgroups of semi-direct products in general, but the situation seemed quite complicated. Would it be easier to find just the structures?

Shaun
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Leppala
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    For each subgroup $L$ of $H$, you have the subgroup $L$ itself, together with $|H:L|$ subgroups isomorphic to $D(L)$. The elements of $D(H) \setminus H$ are all involutions that act by inversion on $H$, so they are all essentially equivalent. Each subgroup isomorphic to $D(L)$ contains exactly $|L|$ of these involutions, which explains why there are $|H:L|$ such subgroups. – Derek Holt Sep 03 '15 at 11:24
  • Ok, thanks for both. A follow-up if that's ok in comments: In addition to the subgroups of $H$ there can be other normal subgroups as well. When $2 \mid |H|$, then we can "separate" the non-trivial center out into a direct factor. For example, in $[24,14]$ we have $G = D(C_2 \times C_6) \cong Z(G) \times S_3$. This direct product gives us some other normal subgroups. Are these two classes all there are? – Leppala Sep 03 '15 at 13:23
  • The normal subgroups $D(L)$ are those with $H^2 \le L$, where $H^2 = {h^2 : h \in H }$. Your remark about the center spltting as a direct factor is not true in general. It is not true in $D(C_4)$ for example. – Derek Holt Sep 03 '15 at 13:55
  • Thanks. My bad, just checked some small examples with GAP and made a hasty guess :) – Leppala Sep 04 '15 at 08:10

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Let $K$ be a subgroup of $D(H)$.

If $K$ is not contained in $H$, then $K$ contains some $g \not\in H$. Then $g$ is an involution and $D(H) = H \rtimes \langle g \rangle$. Hence $K = (H \cap K) \rtimes \langle g \rangle$ and $K \cong D(H \cap K)$.

Mikko Korhonen
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