# Recovering a quotient map of groups from the outer action of the quotient on the kernel

Given any exact sequence of groups $$1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$$

one can define a natural outer action of $C$ on $A$, ie a homomorphism $\rho : C\rightarrow\text{Out}(A)$ given by lifting elements of $C$ to $B$ and restricting the conjugation action to the normal subgroup $A$.

Now forget the above, and suppose we are given groups $A,C$ and a representation $\rho : C\rightarrow\text{Out}(A)$, and further suppose $A$ has trivial center. Does this data determine the group $B$ (up to isomorphism as an extension of $C$ by $A$)?

• Yes I believe so. I seem to remember that, if extensions of this type exist at all, then their equivalence classes are in bijection with $H^3(C,Z(A))$, where the actionof $C$ on $Z(A)$ is induced by the given map $\rho$. I expect you can prove it directly when $Z(A)=1$ by constructing $B$ as a subdirect product of $C$ and the inverse image of $\rho(C)$ in ${\rm Aut}(A)$. – Derek Holt Jul 9 '18 at 7:50
• @Derek Surely the answer is "no"! The given data describes a semidirect product, so to find a counter-example take $B$ to be any group which (a) does not split as a semidirect product, and (b) contains a normal subgroup with trivial centre. (For example, the quaternions satisfy (a) but not (b).) – user1729 Jul 9 '18 at 12:24
• @user1729 Why do you say that the given data describes a semidirect product? Can you suggest a specific counterexample? – Derek Holt Jul 9 '18 at 12:39
• @DerekHolt If we ignore the restriction on $Z(A)$, then the quaternions versus $C_2\times D_2$ works. Writing $Q:=\langle a, i, j, k\mid a^2=1, i^2=j^2=k^2=ijk=a\rangle$, take $A=\langle a\rangle$. Then $i^{-1}ai=iaai=i^2=a$, and similarly for $j$ and $k$. Hence, $\rho: C\rightarrow Out(A)$ is the trivial map, where $C=D_2$ the dihedral group of order $4$. Clearly the direct product $C_2\times D_2$ also satisfies this data. However, $Q\not\cong C_2\times D_2$ (as, for example, $Q$ has elements of order $4$). – user1729 Jul 9 '18 at 12:53
• (The data describes a semidirect product because if $\phi_1Inn(N)=\phi_2Inn(N)$ then $N\rtimes_{\phi_1}H\cong N\rtimes_{\phi_2}H$. That is, it is only equivalence up to outer automorphisms we care about for semidirect products. The data also describes non-split extensions, which is kinda my point.) – user1729 Jul 9 '18 at 12:55

I believe the answer is yes. Suppose that $\rho:C \to {\rm Out}(A)$ is given, and let $\pi: {\rm Aut}(A) \to {\rm Out}(A)$ be the natural map.
Suppose that we are given an extension $1 \to A \to B \overset{\tau}{\to} C \to 1$ as described. We can define a homomorphism $\phi:B \to {\rm Aut}(A) \times C$ by $\phi(b) = (\gamma(b),\tau(b))$, where $\gamma(b)$ is the automorphism of $A$ induced by conjugating by $b$. The fact that $Z(A)=1$ implies that $\phi$ is injective, and its image is precisely $I := \{(\alpha,c) : \pi(\alpha)=\rho(c) \}$. So $B \cong I$, and in fact $B$ is determined up to equivalence of extensions.
• Thank you for this beautiful construction! I had expected a nasty computation with nonabelian cohomology, but this is surprisingly simple. I find it very interesting that we have such a nice description of extensions either when $A$ is centerless (as in your description) or when $A$ is abelian (group cohomology). At first I thought that using this and the upper central series to break a general finite group into abelian and center-free pieces, one could get a "complete" description of group extensions of finite groups, though it seems that doesn't quite work. – user355183 Jul 9 '18 at 22:48
• Btw the motivation from my question comes from "anabelian geometry", where $C$ is the absolute galois group of some field, and $A$ the geometric fundamental group of a hyperbolic curve (which is always center free). In some paper it was stated without proof that the outer representation determines the extension when $A$ is center free, but I couldn't see why, whence this question. – user355183 Jul 9 '18 at 22:50