Schur Multipliers and Universal Central Extensions — Examples

Multiplier of a cyclic group applies to every finite or infinite cyclic group and gives $M(C)=0$.

This is exactly the claimed cyclic-group calculation.

For two abelian groups $B,C$, the alternating universal property gives ${\bigwedge }^2(B\oplus C)\cong {\bigwedge }^{2}B\oplus (B\otimes C)\oplus {\bigwedge }^{2}C$: an alternating pairing on the direct sum is exactly a pairing on each summand plus an arbitrary bilinear cross-pairing. Iterate over the finite summands. Each cyclic summand has zero exterior square, while $C_m\otimes C_n\cong C_{\gcd (m,n)}$ by the relations $m(1\otimes 1)=n(1\otimes 1)=0$.

Applying Multiplier of an abelian group gives the displayed formula, including the empty sum $0$ when $r\le 1$.

$A_5$ has order $60$. Its nonidentity conjugacy classes have sizes $20$ (three-cycles), $15$ (products of two disjoint transpositions), and $12,12$ (the two classes of five-cycles). No proper union of these classes with the identity has order dividing $60$, so $A_5$ is simple. It is nonabelian, hence its commutator subgroup is all of $A_5$: $A_5$ is perfect. Its Sylow-$2$ subgroup has order $4$ and cannot be cyclic, since a permutation of order $4$ in $S_5$ is odd; it is $V=C_2\times C_2$. Its Sylow-$3$ and Sylow-$5$ subgroups are cyclic.

The quotient ${\mathrm{SL}}_2({\mathbb{F}}_5)\to {\mathrm{PSL}}_2({\mathbb{F}}_5)\cong A_5$ has central kernel $\{\pm I\}\cong C_2$. The group ${\mathrm{SL}}_2({\mathbb{F}}_5)$ is perfect. Indeed, elementary row reduction shows it is generated by the elementary matrices $e_{12}(a),e_{21}(a)$. For $d=\mathrm{diag}(2,3)$, conjugation multiplies the upper parameter by $2^2=4$ and the lower parameter by $2^{-2}=4$. Thus $[d,e_{ij}(a)]=e_{ij}(3a)$ for $i\mathrm{\ne }j$; since $3$ is a unit in ${\mathbb{F}}_5$, every elementary generator is a commutator.

The multiplier $M(A_5)$ is finite and annihilated by $60$. For each prime $p\mid 60$, [L1] says that its $p$-primary part is the image of $M(P)$: multiplication by the index $[A_5:P]$, prime to $p$, is an automorphism on the $p$-primary part. The cyclic-group calculation gives $M(P)=0$ for $p=3,5$. For $p=2$, the abelian-multiplier theorem gives $M(V)\cong {\bigwedge }^{2}V\cong C_2$. Consequently $M(A_5)$ is either $0$ or $C_2$.

Choose a free presentation of the perfect group $A_5$. Under Choice, Free-presentation construction is universal gives its universal central extension $u:U\to A_5$, and the kernel corollary identifies $\ker u$ with $M(A_5)$. If this multiplier were zero, universality would give a homomorphic section $A_5\to {\mathrm{SL}}_2({\mathbb{F}}_5)$ of the central quotient. The latter group would then be $C_2\times A_5$, contradicting its perfectness from step 1.2. Hence $M(A_5)\cong C_2$ by step 2.1.

Universality gives a map $U\to {\mathrm{SL}}_2({\mathbb{F}}_5)$ over $A_5$. Its image, together with the central kernel $C_2$, generates ${\mathrm{SL}}_2({\mathbb{F}}_5)$. Because that group is perfect, its commutator subgroup lies in the image, so the map is surjective. Both $U$ and ${\mathrm{SL}}_2({\mathbb{F}}_5)$ have order $2\cdot 60=120$, so it is an isomorphism over $A_5$. This proves the asserted universal central extension.

As $F$ is cyclic, $[F,F]=1$ and $R\cap [F,F]=1$.

Thus Hopf’s quotient is trivial.

Thus $C_2\to D_8\to C_2\times C_2$ is stem.

Let $V=C_2\times C_2$ be its base. If the displayed stem extension $u:D_8\twoheadrightarrow V$ were universal, the maps $x\mapsto (u(x),0)$ and $x\mapsto (u(x),u(x))$ from $D_8$ to the split central extension $V\times V\to V$ would have to be equal. They are distinct because $u$ is surjective and $V\mathrm{\ne }0$. Thus this stem extension is not universal, using only the choice-free uniqueness condition in the definition.

The exterior-square calculation gives $M(C_2\times C_2)\cong C_2$, so both are Schur covers.

$D_8$ has five involutions while $Q_8$ has one, hence they are not isomorphic.