Given a group $Q$ and an abelian group $C$, I want to determine the number $I(Q,C)$ of isomorphism classes of all groups $G$ having a central subgroup $C'$ isomorphic to $C$ such that the quotient $G/C'$ is isomorhic to $Q$. Thus $I(Q,C)$ equals the number of groups (up to isomorphism), that fit into a central extension $$ 1 \to C \to G \to Q \to 1$$ The number of such extensions (up to equivalence) is given by $|H^2(Q,C)|$ (trivial coefficients). This gives an upper bound. But often it is much larger than $I(Q,C)$ (cf. the example below).

Question: Is it possible to obtain the correct value for $I(Q,C)$ out of $H^2(Q,C)$, or at least an improved upper bound ?

Example: Let $C_n$ denote the cyclic group of order $n$. Each $p$-group $G$ of order $n$ has a central subgroup $C_p$. Hence $I(G/C_p,C_p)$ is the number of isomorphism classes of $p$-groups of order $n$.

Take $n=2$. Then there are exactly two groups of order $p^2$, i.e $I(C_p,C_p) =2$, while $|H^2(C_p,C_p)| = |\mathbb{Z}/p\mathbb{Z}| = p$. From these $p$ extensions, $p-1$ belong to the isomorphism class of $C_{p^2}$.

  • $\begingroup$ You have to take into account the size of the outer automorphism group of $C$. Given any extension $Q$ acts on $C$ by outer automorphisms, and this action is trivial if and only if $C$ is central. You probably want to look at the series of papers 'Cohomology theory in abstract groups' by Eilenberg and MacLane. $\endgroup$
    – David Roberts
    Sep 11, 2011 at 22:38

1 Answer 1


Let $Q$ be a group and $A$ a $Q$-module. Call two extensions $G, G'$ of $Q$ by $A$ weakly equivalent, if there is a commutative diagramm $$ 0 \to A \to G \to Q \to 1 $$ $$ \hspace{2pt} \downarrow \hspace{20pt} \downarrow \hspace{20pt} \downarrow $$ $$ 0 \to A \to G' \to Q \to 1 $$ with vertical isomorphisms. Denote the corresponding set of equivalence classes by $W(Q,A)$. Since $G,G'$ are isomorphic, if the extensions are weakly equivalent, $W(Q,A)$ is finer than isomorphism classes, but coarser than $H^2(Q;A)$.

In order to describe the relation between $W(Q,A)$ and $H^2(Q;A)$, some notation is needed: Call $(\varphi, \alpha) \in Aut(Q) \times Aut(A)$ compatible, if $\alpha(\varphi(q)\cdot a) = q \cdot \alpha(a)$ for all $q \in Q, a \in A$. Such a pair induces an automorphism $(\varphi,\alpha)^*$ of $H^2(Q;A)$ (see Brown: Cohomology of Groups, III, after Cor. 8.2).

Taking into account that cohomology is contravariant in the first argument, let $T\subseteq Aut(Q) \times Aut(A)$ be the subgroup of all pairs $(\varphi, \alpha)$ such that $(\varphi^{-1}, \alpha)$ is compatible. Then, $T$ operates on $H^2(Q;A)$ through $(\varphi,\alpha) \cdot x = (\varphi^{-1},\alpha)^*(x)$. Now, the central result is:

There is a bijection between $W(Q,A)$ and the orbits of $H^2(Q;A)$ under the action of $T$.

The proof consists in essential of the fact, that for a 2-cocycle $f: Q\times Q \to A$ and a compatible pair $(\varphi, \alpha)$, the extensions corresponding to $f$ and $f':= \alpha \circ f \circ (\varphi^{-1} \times \varphi^{-1})$ are weakly equivalent.

As noted above, weak equivalence is finer than isomorphism. But in some situations, $W(Q,A)$ will directly classify isomorphism classes.

a) If the center of $Q$ is trivial, then $|W(Q,A)|=I(Q,A)$ (as defined in the question).

b) Suppose $A$ is finite and let the integer $k$ be coprime to $|A|$. Thus, multiplication by $k$ is an automorphism of $A$ that is compatible with $\operatorname{id}_Q$. Hence, for $x \in H^2(Q;A)$, its orbit contains $kx$ for all $k$ comprime to $|A|$. In particular:

If $|H^2(Q;A)|$ is a prime, then $|W(Q,A)| = 2$ and $I(Q,A) \le 2$.

Hence, in your example ahead, there are at most two isomorphism classes of groups of order $p^2$ and since $C_p \times C_p$, $C_{p^2}$ aren't isomorphic, we are done.

Edit: Including a proof of a)

Let $\mathcal{C}$ be the class of groups $G$, satisfying $Z(G) \cong A$ and $G/Z(G) \cong Q$. By fixing such isomorphisms each $G \in \mathcal{C}$ exhibits a central extension $$\mathcal{E}_G: \quad\quad A \cong Z(G) \hookrightarrow G \twoheadrightarrow G/Z(G) \cong Q.$$ The first key observation is: An isomorphism $\phi: G \to H$ maps $Z(G)$ isomorphically onto $Z(H)$ and therefore induces a commutative diagramm with vertical isomorphisms: $$\mathcal{E}_G: \quad\quad A \cong Z(G) \hookrightarrow G \twoheadrightarrow G/Z(G) \cong Q$$ $$\hspace{17pt} \phi \downarrow \hspace{17pt} \phi \downarrow \hspace{17pt} \bar{\phi} \downarrow $$ $$\mathcal{E}_H: \quad\quad A \cong Z(H) \hookrightarrow H \twoheadrightarrow H/Z(H) \cong Q$$ Hence $\mathcal{E}_G$ and $\mathcal{E}_H$ are weakly equivalent and we obtain a map $$\mathcal{C}/\cong \to \mathcal{W}(Q,A),\quad [G] \mapsto [\mathcal{E}_G].$$ Since a weak equivalence between $\mathcal{E}_G$ and $\mathcal{E}_H$ implies $G \cong H$, this map is injective. Surjectivity follows from the second key observation: Let $$\mathcal{E}: \quad\quad A \hookrightarrow G \overset{\kappa}{\twoheadrightarrow} Q$$ be a central extension, i.e. $A \le Z(G)$. Since $\kappa$ is epi, $\kappa(Z(G)) \le Z(Q) = 1$, implying $Z(G) = A$. Thus $G \in \mathcal{C}$ and $[\mathcal{E}_G] = [\mathcal{E}]$.

  • $\begingroup$ Ralph, thank you very much for your answer. In fact, considering those orbits drastically reduces the number of extensions that can arise from an isomorphism class. Unfortunately I don't understand, why in case of $Z(Q) =1$, the orbits correspond 1-1 to isomorphism classes. Can you give a hint about how to prove this ? $\endgroup$ Sep 22, 2011 at 19:41
  • $\begingroup$ I'm sorry for responding that late. $\endgroup$
    – Ralph
    Oct 6, 2011 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.