Group extension

A group extension of $G$ by an abelian $G$-module $A$ is a short exact sequencewhose conjugation action on $A$ is the specified $G$-action.

For a specified finite group action on $A={\mathbb{Z}}^n$, extensions are classified by $e\in H^2(Q,A)$. For an element $q$ of prime order $p$, a lift $g$ satisfies $g^p=a\in A^q$; changing the lift changes $a$ by $N_{q}v=(1+q+\cdots +q^{p-1})v$. Thus the extension is torsion-free precisely when every restriction to a prime-order cyclic subgroup is nonzero in $A^q/N_{q}A$. Any finite-order element has a prime-order power, so this tests all torsion. An arbitrary finite quotient must not be confused with faithful holonomy on a maximal translation lattice.

A group extension is split when its quotient homomorphism admits a group-homomorphic section. Equivalently, the middle group is a semidirect product of the kernel by the quotient.

If a group homomorphism section $i:G/N\to G$ satisfies $\pi i=\mathrm{id}$, then each $g\in G$ has a unique form $i(q)n$ with $q\in G/N$ and $n\in N$: take $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Thus $(q,n)\mapsto i(q)n$ is a bijection. It is a homomorphism for the ordinary direct product of groups exactly when the section image centralizes $N$. In general it describes a semidirect product with a nontrivial conjugation action.

Two extensions of $G$ by the same $G$-module $M$ are equivalent when an isomorphism between their middle groups commutes with the inclusions of $M$ and the projections to $G$. Splitness is preserved under this equivalence.

For homomorphisms $E\to H$ and $G\to H$, their fiber product isIt is a subgroup of $E\times G$. Pulling a group extension back along $G\to H$ uses this fiber product as its middle group.

For a finite group presentation $H=⟨x_1,\dots ,x_m\mid r_1,\dots ,r_n⟩$, let $F$ be the free group on its generators and $\pi :F\to H$. The corresponding fiber product of groups is generated by $(x_i,x_i)$ and $(1,r_j)$. Conjugation by the diagonal pairs supplies every pair $(1,r)$ with $r$ in the normal closure of the relators, and $(u,v)=(u,u)(1,u^{-1}v)$ proves the generating claim. Membership of $(1,w)$ is equivalent to $w=1$ in $H$, so this finitely generated subgroup can have undecidable membership even though $F\times F$ has soluble word problem.

Equivalence classes of group extensions of $G$ by an abelian $G$-module $A$ correspond to $H^2(G,A)$. A section $s:G\to E$ produces the extension cocycleand changing the section changes $c$ by a group coboundary.

With a fixed identification of the kernel with $\mathbb{Z}$ and quotient with ${\mathbb{Z}}^2$, these central group extensions are classified by $n\in \mathbb{Z}$. Choose lifts $x,y$ of the quotient generators; their commutator is $z^n$ and is unchanged by central changes of lifts. Every element has unique normal form $z^m{y}^b{x}^a$, proving this invariant is complete. A cocycle representative is $c((a,b),(c,d))=nad$, whose cocycle identity follows from bilinearity. Its extension multiplication is $(a,b,m)(c,d,l)=(a+c,b+d,m+l+nad)$. The zero parameter gives ${\mathbb{Z}}^3$ and parameter one gives the integer Heisenberg group. The Koszul resolution for a rank-two free abelian group independently computes $H^2({\mathbb{Z}}^2,\mathbb{Z})\cong \mathbb{Z}$.

An extension cocycle is the two-cocycle obtained from a section of a group extension. It measures the failure of the section to be a group homomorphism.