Schur Multipliers and Universal Central Extensions

Use one generator $x_g$ for each $g\in G$ and the finitely many relations $x_g{x}_h=x_{gh}$ and $x_1=1$.

Every word reduces to one $x_g$, so the presented group maps bijectively to $G$ and is a finite presentation.

Normality of $R$ gives $[F,R]\le R$, and the commutator identities make $[F,R]$ normal in $F$.

Every generator $[f,r]$ is in $[F,F]$, so $[F,R]\le R\cap [F,F]$ and the quotient exists.

The hypotheses and group-homology conventions are exactly those of [L1]. Its first arrow is the free-presentation transgression, and its remaining arrows come from the inclusion $R\hookrightarrow F$ and quotient $F\twoheadrightarrow G$.

The exactness and initial injection asserted by [L1] therefore give the displayed sequence under the retained dependent-choice and resolution hypotheses.

Under the stated hypotheses, Low-degree sequence of a free presentation gives an injection $H_2(G;\mathbb{Z})\to R\mathrm{/}[F,R]$ whose image is the kernel of $R\mathrm{/}[F,R]\to F_{\mathrm{a}\mathrm{b}}$. This kernel is $(R\cap [F,F])\mathrm{/}[F,R]$, since the map sends $r[F,R]$ to $r[F,F]$.

Exactness identifies that kernel with $H_2(G;\mathbb{Z})=M(G)$ by Schur multiplier, giving the stated isomorphism.

For each presentation, the conditional five-term sequence injects $H_2(G;\mathbb{Z})$ into $R\mathrm{/}[F,R]$ with image the kernel of $R\mathrm{/}[F,R]\to F_{\mathrm{a}\mathrm{b}}$. That kernel is $(R\cap [F,F])\mathrm{/}[F,R]$, since the map sends $r[F,R]$ to $r[F,F]$. By the definition of the Schur multiplier this gives $(R\cap [F,F])\mathrm{/}[F,R]\cong M(G)$, and the same argument gives $(R^{\mathrm{\prime }}\cap [F^{\mathrm{\prime }},F^{\mathrm{\prime }}])\mathrm{/}[F^{\mathrm{\prime }},R^{\mathrm{\prime }}]\cong M(G)$.

Composing the first isomorphism with the inverse of the second identifies the two presentation quotients. Thus their isomorphism type depends only on $G$, through $M(G)$.

Modulo $[F,R]$, every conjugate of a defining relator has the same class as that relator. Since $R$ is the normal closure of the finite relator set, their images generate $R\mathrm{/}[F,R]$. This quotient is abelian because $[R,R]\le [F,R]$.

Hopf's formula, under the stated hypotheses, identifies $M(G)$ with the subgroup $(R\cap [F,F])\mathrm{/}[F,R]$ of this finitely generated abelian group. Subgroups of finitely generated abelian groups are finitely generated, so $M(G)$ is finitely generated.

Hopf’s numerator is $1\cap [F,F]=1$ and its denominator is $[F,1]=1$.

Thus the Hopf quotient, and hence $M(F)$, is zero.

The group $F$ is abelian, hence $R\cap [F,F]=1$.

Under the stated hypotheses, Hopf's formula gives $M(C)=(R\cap [F,F])\mathrm{/}[F,R]=0$ in both cases. The trivial group is also cyclic; its free presentation $F=R=1$ gives the same result.

The tensor universal property gives a unique map $A\otimes A\to B$ carrying $a\otimes a^{\mathrm{\prime }}$ to $b(a,a^{\mathrm{\prime }})$.

Because $b(a,a)=0$, it kills the defining subgroup and factors uniquely through ${\bigwedge }^{2}A$.

In $F\mathrm{/}[F,R]$, the subgroup $R\mathrm{/}[F,R]$ is central. Since $[F,F]\le R$, this quotient has nilpotency class at most two. Thus its commutator is alternating and bilinear. Changing either lift by an element of $R$ does not change its commutator, so the rule $\bar{x}\wedge \bar{y}\mapsto [x,y][F,R]$ defines a surjective homomorphism $\mathrm{\Phi }:{\bigwedge }^{2}A\to [F,F]\mathrm{/}[F,R]$.

To check that there are no further relations, write $X=F_{\mathrm{a}\mathrm{b}}$. The class-two quotient $F\mathrm{/}[F,[F,F]]$ is the free class-two group on the chosen free basis. Its central commutator subgroup is ${\bigwedge }^{2}X$: ordering the basis, every commutator word reduces using $[xy,z]=[x,z][y,z]$ and $[x,yz]=[x,y][x,z]$ to a product of basic commutators $[x_i,x_j]$ with $i<j$. Their independence follows by mapping the basis into the explicit class-two group $(X\oplus {\bigwedge }^{2}X,\cdot )$ with $(u,a)\cdot (v,b)=(u+v,a+b+{\sum }_{i>j}{u}_i{v}_j(e_i\wedge e_j))$; each word has finite support, so the displayed sum is finite.

The image of $[F,R]$ in ${\bigwedge }^{2}X$ is generated by $x\wedge r$ with $x\in X$ and $r$ in the image $K$ of $R$ in $X$. The alternating universal property identifies $({\bigwedge }^{2}X)\mathrm{/}(X\wedge K)$ with ${\bigwedge }^2(X\mathrm{/}K)={\bigwedge }^{2}A$. Hence $\mathrm{\Phi }$ is injective as well as surjective. Since $[F,F]\le R$, Hopf formula for the Schur multiplier identifies $M(A)=(R\cap [F,F])\mathrm{/}[F,R]=[F,F]\mathrm{/}[F,R]$ under the stated hypotheses. Presentation maps preserve commutators and exterior products, and the canonical Hopf identification is natural, so the resulting isomorphism is natural in $A$.

Initiality produces unique maps $U\to V$ and $V\to U$ over $G$.

Their composites are the unique endomorphisms over $G$, hence identities; the maps are inverse and unique.

For every abelian group $A$ and homomorphism $\varphi :G\to A$, the two maps $x\mapsto (u(x),0)$ and $x\mapsto (u(x),\varphi (u(x)))$ from $U$ to the split central extension $G\times A\to G$ must agree by universality. Since $u$ is surjective, $\varphi =0$. Taking $A=G_{\mathrm{a}\mathrm{b}}$ and $\varphi$ the quotient map gives $G_{\mathrm{a}\mathrm{b}}=0$, so $G$ is perfect.

Conversely, let $G=F\mathrm{/}R$ be perfect, using for example the free group on the underlying set of $G$. Under Choice, Free-presentation construction is universal proves that $[F,F]\mathrm{/}[F,R]\to G$ is a universal central extension. Its proof identifies the exact use of Choice: lifting every free generator through each central quotient.

Steps 1.1 and 1.2 prove the two implications.

Perfectness gives $F=[F,F]R$, so $[F,F]\mathrm{/}[F,R]\to G$ is onto and has kernel $(R\cap [F,F])\mathrm{/}[F,R]$.

The latter kernel commutes with $[F,F]\mathrm{/}[F,R]$, since every $[r,f]$ is killed; therefore the extension is central.

The chosen basis lifts extend uniquely to a homomorphism $h:F\to E$ over $G$. Every $r\in R$ maps into the central kernel of $E\to G$, so $h([F,R])=1$. Restricting $h$ to $[F,F]$ therefore gives a homomorphism $[F,F]\mathrm{/}[F,R]\to E$ over $G$. Since $G$ is perfect, the displayed source maps onto $G$, and the preceding central-extension lemma gives its central kernel.

If $v,w:[F,F]\mathrm{/}[F,R]\to E$ are two maps over $G$, their pointwise ratio $v(x)w(x{)}^{-1}$ lies in the central kernel. Centrality makes this ratio a homomorphism to an abelian group. The source is perfect: write elements $a,b\in F$ as $a=a^{\mathrm{\prime }}r$, $b=b^{\mathrm{\prime }}s$ with $a^{\mathrm{\prime }},b^{\mathrm{\prime }}\in [F,F]$ and $r,s\in R$, which is possible because $G$ is perfect; modulo $[F,R]$ one has $[a,b]=[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]$. Its generators are thus commutators inside the source. Every homomorphism from it to an abelian group is zero, so $v=w$. Existence from step 1.1 and this uniqueness prove universality.

Its kernel is $(R\cap [F,F])\mathrm{/}[F,R]$.

Under the stated hypotheses, the free-presentation five-term exact sequence identifies its initial $H_2(G;\mathbb{Z})$ with the kernel of $R\mathrm{/}[F,R]\to F_{\mathrm{a}\mathrm{b}}$, namely $(R\cap [F,F])\mathrm{/}[F,R]$. This is $M(G)$ by definition.

For an abelian group $A$ and a homomorphism $\varphi :U\to A$, the maps $x\mapsto (u(x),0)$ and $x\mapsto (u(x),\varphi (x))$ from $U$ to the split central extension $G\times A\to G$ are both over $G$. Universality makes them equal, so every such $\varphi$ vanishes. Taking $A=U_{\mathrm{a}\mathrm{b}}$ shows that $U$ is perfect.

Let $e:E\twoheadrightarrow U$ be any central extension. The composite $[E,E]\to U\to G$ is surjective because $e([E,E])=[U,U]=U$. If $x\in [E,E]$ maps to $1$ in $G$, then $e(x)\in \ker u\subseteq Z(U)$, whence $[x,E]\subseteq \ker e\subseteq Z(E)$ and therefore $[x,[E,E]]=1$. Thus $[E,E]\to G$ is a central extension.

Universality of $u$ gives a map $s:U\to [E,E]$ over $G$. Both $e\circ s$ and ${\mathrm{id}}_U$ are maps from $U$ to the central extension $u:U\to G$ over $G$, so uniqueness gives $e\circ s={\mathrm{id}}_U$. Consequently every central extension of $U$ splits.

Since $U$ is perfect, the free-presentation theorem under Choice gives a universal central extension $v:V\to U$. Under the resolution hypotheses, its kernel is $M(U)$ by Kernel of the universal central extension. Step 3.1 gives a homomorphic section, so centrality yields $V\cong M(U)\times U$. The argument of step 1.1 applied to $v$ makes $V$ perfect. Abelianizing this product gives $M(U)=0$. Also $U_{\mathrm{a}\mathrm{b}}=0$ by step 1.1, so $H_1(U;\mathbb{Z})=H_2(U;\mathbb{Z})=0$ and $U$ is superperfect.

Each $C_n$ is free abelian on $G^n$. Because $A$ has trivial $G$-action, the tensor-Hom adjunction gives a natural isomorphism ${\mathrm{Hom}}_{\mathbb{Z}}(C_{\bullet },A)\cong {\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },A)$. The latter cochain complex computes $H^{\ast }(G;A)$ by The bar cochain complex computes group cohomology, while $H_{\ast }(C_{\bullet })$ computes integral group homology under the supplied resolution data. Choice permits [L1] to be applied to the free integral complex $C_{\bullet }$. In degree two it gives $0\to {\mathrm{Ext}}_{\mathbb{Z}}^1(H_1(G;\mathbb{Z}),A)\to H^2(G;A)\to {\mathrm{Hom}}_{\mathbb{Z}}(H_2(G;\mathbb{Z}),A)\to 0$.

Since $H_1(G;\mathbb{Z})=G_{\mathrm{a}\mathrm{b}}$ and $H_2(G;\mathbb{Z})=M(G)$, this is the displayed sequence. The splitting qualification follows from [L2].

Perfectness gives $G_{\mathrm{a}\mathrm{b}}=0$, hence ${\mathrm{Ext}}_{\mathbb{Z}}^1(G_{\mathrm{a}\mathrm{b}},A)=0$. The degree-two universal-coefficient sequence therefore identifies $H^2(G;A)$ naturally with $\mathrm{Hom}(M(G),A)$.

The extension-classification theorem identifies $H^2(G;A)$ with equivalence classes of extensions inducing the trivial action, namely central extensions. Composing the two bijections proves the claim.

The cited theorem of Weibel states that, for a finite group $G$, multiplication by $\mathrm{\mid }G\mathrm{\mid }$ annihilates $H_n(G;A)$ for every $n>0$ and every $G$-module $A$. Apply it to the trivial module $A=\mathbb{Z}$.

Thus multiplication by $\mathrm{\mid }G\mathrm{\mid }$ is zero on $H_n(G;\mathbb{Z})$, as claimed.

The finite-presentation lemma gives a finite presentation of $G$; the finitely-presented-multiplier corollary makes $M(G)$ finitely generated under the stated hypotheses. The finite-group homology torsion lemma says $\mathrm{\mid }G\mathrm{\mid }$ annihilates this positive-degree integral homology group.

A finitely generated abelian group of bounded exponent is finite, so $M(G)$ is finite.

The quotient $A\mathrm{/}M\cong R\mathrm{/}(R\cap [F,F])$ embeds in the finite-rank free abelian group $F_{\mathrm{a}\mathrm{b}}$, so it is free abelian. Thus $0\to M\to A\to A\mathrm{/}M\to 0$ splits. Let $S\mathrm{/}[F,R]$ be the image of a splitting, a complement to $M$ in $A$. Because $[F,R]\le S\le R$, conjugation by any $f\in F$ preserves $S$, so $S$ is normal in $F$.

The extension $R\mathrm{/}S\to F\mathrm{/}S\to G$ is central because $[F,R]\le S$. Projection $A\to A\mathrm{/}(S\mathrm{/}[F,R])$ restricts to an isomorphism from $M$, so $R\mathrm{/}S\cong M\cong M(G)$ by Hopf's formula. Since every element of $M$ comes from $R\cap [F,F]$, the kernel $R\mathrm{/}S$ lies in $[F\mathrm{/}S,F\mathrm{/}S]$. Thus it is a stem extension with multiplier kernel. The kernel is finite by Multiplier of a finite group is finite, hence $F\mathrm{/}S$ is finite because $G$ is finite. This is a Schur cover.

Schur multiplier defines $M(G)=H_2(G;\mathbb{Z})$. The group $H^2(G;{\mathbb{C}}^{\times })$ is a degree-two cohomology group with different coefficients and is not the specified definition. No comparison theorem is needed to refute a claim about which definition this library uses.

It is therefore false to present $H^2(G;{\mathbb{C}}^{\times })$ as this library's definition of the multiplier.

In the first presentation $F=R=\mathbb{Z}$, so the numerator $R\cap [F,F]=[\mathbb{Z},\mathbb{Z}]$ is trivial.

In the second presentation $F=R=F(a,b)$, so the numerator is $[F(a,b),F(a,b)]$. It contains the commutator $ab{a}^{-1}{b}^{-1}$, which is a nonempty reduced word and hence nonidentity by [F2]. The two numerator groups are therefore not isomorphic.

The denominators are respectively trivial and $[F(a,b),F(a,b)]$, so both Hopf quotients are trivial. This agrees with the qualified invariant comparison in [L1]; it does not make their numerators isomorphic.

Suppose a universal central extension $u:U\twoheadrightarrow C_2$ exists. The split extension $C_2\times C_2\to C_2$ is central. There are two maps over $C_2$ from $U$ to it: $x\mapsto (u(x),0)$ and $x\mapsto (u(x),u(x))$. They are distinct because $u$ is surjective and $C_2$ is nontrivial.

This contradicts the uniqueness required of a universal central extension. Thus $C_2$ is a counterexample to the stated assertion, independently of any choice premise.

Its kernel is central, but the total group is abelian, so its commutator subgroup is trivial and does not contain the nontrivial kernel.

The extension is central but not stem, refuting the assertion.

The abelian-multiplier theorem gives $M(V)\cong {\bigwedge }^{2}V\cong C_2$. In $D_8=⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, the center and commutator subgroup are both $⟨r^2⟩\cong C_2$, and the quotient is $V$. In $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$, the center and commutator subgroup are both $\{\pm 1\}\cong C_2$, and its quotient is also $V$. Both quotient maps are therefore finite stem extensions with multiplier kernel and hence Schur covers.

The group $D_8$ has five nonidentity involutions whereas $Q_8$ has one, so the two covers are not isomorphic. The stated uniqueness claim is false.

Suppose there are splitting homomorphisms $s_G:\mathrm{Hom}(M(G),A)\to H^2(G;A)$ natural in $G$, with $p_G{s}_G=\mathrm{i}\mathrm{d}$. For every subgroup inclusion $i:L\hookrightarrow V$ with $L\cong C_2$, [L2] gives $M(L)=0$, so naturality forces $i^{\ast }{s}_V(\lambda )=s_L(\lambda \circ M(i))=s_L(0)=0$ for every $\lambda :M(V)\to A$.

The map $b((x_1,x_2),(y_1,y_2))=x_1{y}_2+x_2{y}_1$ is alternating and bilinear to $A$. It takes value $1$ on the standard basis pair, so [L3] supplies a nonzero homomorphism $\lambda :M(V)\to A$. Put $\alpha =s_V(\lambda )$. Then $p_V(\alpha )=\lambda \mathrm{\ne }0$.

Represent $\alpha$ by a central extension $1\to A\to E\stackrel{\pi }{\to }V\to 1$. Its restriction to each of the three order-two subgroups $L\le V$ is zero by step 1.1. Hence each preimage ${\pi }^{-1}(L)$ is a split central extension, isomorphic to $C_2\times C_2$, and every element of that preimage has square $1$.

Every element of $E$ either belongs to the kernel $A$, or maps to a nonzero vector of $V$ and therefore belongs to one of these three preimages. Thus every element of $E$ has square $1$. For $x,y\in E$, this gives $xy=(xy{)}^{-1}=y^{-1}{x}^{-1}=yx$, so $E$ is abelian. Choose lifts $u,v$ of the two standard basis vectors of $V$. Since $u^2=v^2=1$ and $uv=vu$, the map $(a,b)\mapsto u^a{v}^b$ is a homomorphic section of $\pi$.

This section makes $\alpha =0$, contradicting $p_V(\alpha )=\lambda \mathrm{\ne }0$. Therefore no splitting can be natural in the group variable even for the fixed coefficient group $C_2$, and in particular none is natural in both variables. Under the Axiom of Choice, the individual sequences split after choices by the separate splitting theorem cited in [L1].