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 '11 at 22:38

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 ? $\endgroup$ – Todd Leason Sep 22 '11 at 19:41
  • $\begingroup$ I'm sorry for responding that late. $\endgroup$ – Ralph Oct 6 '11 at 11:21

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