Suppose I have groups $A,B$ and $C$ for which the following information is given:

1) The group $G_{AB}$ is a central extension of $B$ by $A$, where the abelian group $B$ acts trivially

2) The group $G_{BC}$ is an extension of $C$ by $B$, where $C$ does not act trivially, and the action is given by $\theta: C \rightarrow Aut(B)$.

Then, under what conditions can we make the following statement?

There exists a group $G_{ABC}$ which is an extension of $G_{BC}$ by $A$ as well as an extension of $C$ by $G_{AB}$, where the respective actions are induced by the actions given in (1) and (2).

In other words, the following diagram needs to be filled in by groups such that every row and column is a group extension (thanks to Neil Strickland for making this precise): $\require{AMScd}$ \begin{CD} A @>>> G_{AB} @>>> B \\ @| @VVV @VVV \\ A @>>> G_{ABC} @>>> G_{BC} \\ & @VVV @VVV \\ && C @= C \end{CD}

Furthermore, when is $G_{ABC}$ unique?

  • $\begingroup$ In an extension of $B$ by $A$, it's not clear what you mean by "$B$ acts trivially"; this makes sense in a split extension; in general you have a homomorphism $B\to\mathrm{Out}(A)$, which you can ask to be trivial. $\endgroup$
    – YCor
    Jul 12, 2018 at 12:59
  • $\begingroup$ @YCor The extension $G_{AB}$ is a central extension- I have edited the question. Does this help? $\endgroup$ Jul 12, 2018 at 13:38
  • $\begingroup$ Yes, "central extension" makes sense. $\endgroup$
    – YCor
    Jul 12, 2018 at 13:39
  • $\begingroup$ Saying the action is given by $\theta:C \to {\rm Aut}(B)$ in 2) does not make sense unless $B$ is abelian. Perhaps you mean $\theta:C \to {\rm Out}(B)$ ? $\endgroup$
    – Derek Holt
    Jul 12, 2018 at 14:01
  • 1
    $\begingroup$ Normally you would expect to supply an action of $C$ on $A$ that was compatible with its action on $B$. If extensions exist, then the equivalence classes are bijective with $H^2(C,A)$. One situation in which there is a unique such extension is when $A$ and $C$ are finite with coprime orders. $\endgroup$
    – Derek Holt
    Jul 12, 2018 at 14:56

1 Answer 1


This situation was studied by Eilenberg and MacLane in a series of papers in the 1940s. They studied exact sequences $$1 \to A \to G_{AB} \stackrel{\alpha}{\to} G_{BC} \stackrel{\beta}{\to} C \to 1$$ in which $A$ maps into the centre of $G_{AB}$, where we are given an action of $G_{BC}$ on $G_{AB}$ that restricts to the conjugation action on $B = {\rm Im}(\alpha)=\ker(\beta)$. So we get an induced action of $C$ on $A$. This is called a crossed sequence.

In S. Eilenberg and S. MacLane, Cohomology theory in abstract groups II, Ann. of Math. 48 (1947), 326-421 (for example), an equivalence relation is defined on such sequences, and it is shown that the equivalence classes correspond to elements of $H^3(C,A)$. Furthermore, the sequence is derived from an extension $G_{ABC}$ if and only if it corresponds to the zero element of $H^3(G,A)$. In that case, the equivalence classes of extensions $G_{ABC}$ are in one-one correspondence with $H^2(C,A)$. This is proved in Theorem 11.1 of that paper.


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