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Let $H$ be a group, $\phi$ an automorphism of $H$ of order n and fix $h_0 \in H$. I wonder, what the restrictions are, such that $$G:= \lt H,g \mid g^n=h_0,\quad \forall h \in H: ghg^{-1}=\phi(h) \gt$$ defines a group which has $H$ as normal subgroup such that $G/H$ is cyclic of order n.

There are two obvious restrictions:

(1) $h_0$ has to be in the center of $H$. For, $h_0hh_0^{-1} = g^nhg^{-n}=\phi^n(h) = h$, since $\phi$ has order n.

(2) $\phi(h_0) = h_0$. For, $\phi(h_0) = gh_0g^{-1}=gg^ng^{-1} = g^n = h_0$.

But I can't figure out, if there some more restrictions.

I tried to apply the classification of extensions with non-abelian kernel (Kenneth Brown, Cohomology of Groups, chapter IV, §6), but that requires to consider $H^3(Out(H),C)$ ($C$ the center of $H$) and I'm unable to do so, because I have no information about $Out(H)$ and $C$.

Any help is appreciated.

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Your conditions are sufficient. Indeed, consider the $H$-by-cyclic group $G_0=\langle H, g\mid ghg^{-1}=\phi(h), h\in H\rangle$. By (1) and (2), both $g^n$ and $h_0$ are central in that group. Hence the element $u=h_0^{-1}g^n$ is central. Now factor out the central subgroup $\langle u\rangle$: $G=G_0/\langle u\rangle$. That group $G$ is what you need.

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  • $\begingroup$ Ah, I see, $G$ can be realized as central quotient of the semi-direct product $G_0=H \times_\varphi \mathbb{Z}$ where $\varphi: \mathbb{Z} \to Aut(H), k \mapsto \phi^k$. That's good. Thanks a lot, Mark. $\endgroup$
    – Todd Leason
    Commented Sep 20, 2011 at 23:36

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