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 Lyndon--Hochschild--Serre low-degree homology sequence for this extension is $H_2(F;\mathbb{Z})\to H_2(G;\mathbb{Z})\to R\mathrm{/}[F,R]\to F_{\mathrm{a}\mathrm{b}}\to G_{\mathrm{a}\mathrm{b}}\to 0$. This is the five-term sequence recorded in the cited source.

A free group has zero second integral homology, so the first map has zero source and exactness gives the displayed sequence beginning with $0$.

In the low-degree exact sequence, the kernel of $R\mathrm{/}[F,R]\to F_{ab}$ is $(R\cap [F,F])\mathrm{/}[F,R]$.

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

Hopf's theorem gives isomorphisms $(R\cap [F,F])\mathrm{/}[F,R]\cong M(G)$ and $(R^{\mathrm{\prime }}\cap [F^{\mathrm{\prime }},F^{\mathrm{\prime }}])\mathrm{/}[F^{\mathrm{\prime }},R^{\mathrm{\prime }}]\cong M(G)$.

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. Hence the finitely many defining relators generate the abelian group $R\mathrm{/}[F,R]$.

Hopf's formula 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 $R\cap [F,F]=1$.

Hopf’s formula gives $M(C_n)=0$, and the free case gives the infinite cyclic 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 ${\bigwedge }^{2}A$.

The rule $\bar{x}\wedge \bar{y}\mapsto [x,y][F,R]$ is independent of the chosen lifts: changing a lift by an element of $R$ changes the commutator by an element of $[F,R]$. It is alternating and bilinear modulo $[F,R]$, so the universal property gives a homomorphism ${\bigwedge }^{2}A\to [F,F]\mathrm{/}[F,R]$.

Conversely, the commutator quotient $[F,F]\mathrm{/}[F,R]$ is generated by the classes of $[x,y]$, subject exactly to the alternating bilinear commutator relations; sending such a class to $\bar{x}\wedge \bar{y}$ is therefore a well-defined inverse. Hopf's formula gives $M(A)=[F,F]\mathrm{/}[F,R]\cong {\bigwedge }^{2}A$, and the construction 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. Then $F=[F,F]R$, so $[F,F]\mathrm{/}[F,R]\to G$ is a central surjection. Given any central extension $E\to G$, lift the free generators of $F$ to $E$. The resulting map $F\to E$ kills $[F,R]$ on $[F,F]$ because the kernel of $E\to G$ is central. Different choices of lifts differ by central kernel elements and hence agree on $[F,F]$, giving a canonical map $[F,F]\mathrm{/}[F,R]\to E$ over $G$. Writing $F^{\mathrm{\prime }}=[F,F]$, the identity $F=F^{\mathrm{\prime }}R$ implies $F^{\mathrm{\prime }}=[F^{\mathrm{\prime }},F^{\mathrm{\prime }}][F,R]$, so $[F,F]\mathrm{/}[F,R]$ is perfect. The pointwise difference of any two maps from this group to $E$ over $G$ is therefore a homomorphism to the central abelian kernel of $E\to G$, and must vanish. Thus the canonical map is unique, proving universality.

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 induced map from $F$ kills $[F,R]$ on commutators because the kernel of $E\to G$ is central. It therefore restricts to a map $[F,F]\mathrm{/}[F,R]\to E$ over $G$.

Two choices of lifts of the free generators differ by elements of the central kernel, so their maps agree on every commutator and hence on $[F,F]$. Thus the descended map is independent of the lifts and is the unique map over $G$, proving the universal property.

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

Hopf’s formula identifies this kernel with $M(G)$.

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 gives a universal central extension $v:V\to U$, with kernel $M(U)$ by Kernel of the universal central extension. Step 3.1 splits it, so $V\cong M(U)\times U$. The argument of step 1.1, applied to $v$, also makes $V$ perfect. Abelianizing the displayed product therefore gives $M(U)=0$. Hence $H_1(U;\mathbb{Z})=H_2(U;\mathbb{Z})=0$, and Superperfect group makes $U$ superperfect.

The resulting chain complex is degreewise free over $\mathbb{Z}$, so the cohomological universal-coefficient theorem in degree two gives $0\to {\mathrm{Ext}}_{\mathbb{Z}}^1(H_1(G;\mathbb{Z}),A)\to H^2(G;A)\to {\mathrm{Hom}}_{\mathbb{Z}}(H_2(G;\mathbb{Z}),A)\to 0$ naturally.

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.

Its multiplier is finitely generated, and $\mathrm{\mid }G\mathrm{\mid }$ annihilates it because it is positive-degree integral homology.

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 free abelian group $F_{\mathrm{a}\mathrm{b}}$, so it is free abelian. Hence $0\to M\to A\to A\mathrm{/}M\to 0$ splits. Choose a complement $S\mathrm{/}[F,R]$ to $M$ in $A$.

The extension $R\mathrm{/}S\to F\mathrm{/}S\to G$ is central because $[F,R]\le S$. Its kernel is $R\mathrm{/}S\cong M(G)$ by Hopf's formula, and it lies in $[F\mathrm{/}S,F\mathrm{/}S]$ because the chosen complement meets $M$ trivially. Thus it is a stem extension with multiplier kernel. The kernel and $G$ are finite, so $F\mathrm{/}S$ is finite, and Schur covering group makes it a Schur cover.

That definition is $M(G)=H_2(G;\mathbb{Z})$ for every group. The cohomological group $H^2(G;{\mathbb{C}}^{\times })$ is a different construction. Indeed, because ${\mathbb{C}}^{\times }$ is divisible, ${\mathrm{Ext}}_{\mathbb{Z}}^1(G_{\mathrm{a}\mathrm{b}},{\mathbb{C}}^{\times })=0$, so the universal-coefficient sequence identifies it with the character dual $\mathrm{Hom}(M(G),{\mathbb{C}}^{\times })$ for every $G$ (with trivial coefficients), not with $M(G)$ itself.

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

Their groups $(R\cap [F,F])\mathrm{/}[F,R]$ are built from different free groups and there is no presentation-free identification between the displayed quotients from their formulas alone.

Independence is obtained only after Hopf's theorem identifies each quotient with the invariant $M(G)$, as in Hopf formula is presentation-independent. It is not a formal or "obvious" consequence of writing the quotient.

Its commutator subgroup is trivial, so $C_2$ is not perfect.

By Existence criterion for universal central extensions, $C_2$ has no universal central extension.

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.

Since $M(V)\cong {\bigwedge }^{2}V\cong C_2$, both $C_2\to D_8\to V$ and $C_2\to Q_8\to V$ are stem extensions with multiplier kernel. Thus both are Schur covers of $V$.

The group $D_8$ has five involutions whereas $Q_8$ has one, so the two covers are not isomorphic. Schur covers are therefore not unique in general.

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. Individual sequences still split after choices by [L1].