= Solution
The <Schur multiplier> of a <group> $G$ is
$$
M(G)=H_2(G,\mathbb Z),
$$
the second <group homology> group with trivial integral coefficients. If $G\cong F/R$ is a <free presentation>, <Hopf's formula> states that
$$
M(G)\cong\frac{R\cap[F,F]}{[F,R]}.
$$
Write $I_F=\ker(\mathbb ZF\to\mathbb Z)$ for the <augmentation ideal>. The <presentation relation sequence> is
$$
0\longrightarrow R/[R,R]
\longrightarrow\mathbb ZG\otimes_{\mathbb ZF}I_F
\longrightarrow\mathbb ZG
\longrightarrow\mathbb Z\longrightarrow0,
$$
where $r[R,R]\mapsto1\otimes(r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb ZF$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb Z$ are free $\mathbb ZG$-modules. Resolving the <relation module> $R/[R,R]$ by free modules and splicing produces a <free resolution> of $\mathbb Z$.
Apply the right-exact functor $\mathbb Z\otimes_{\mathbb ZG}-$ to this partial resolution. Its degree-two <homology> is the kernel of
$$
(R/[R,R])_G\longrightarrow
\left(\mathbb ZG\otimes_{\mathbb ZF}I_F\right)_G.
$$
The <coinvariant module> on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the <abelianization> $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel is
$$
\ker\left(R/[F,R]\longrightarrow F/[F,F]\right)
=\frac{R\cap[F,F]}{[F,R]}.
$$
This proves <Hopf's formula>.
For an <abelian group> $A$, the <Schur multiplier of an abelian group> is $M(A)\cong\bigwedge^2A$. One way to see the direct-sum rule is the degree-two <Künneth theorem>:
$$
H_2(A\times B,\mathbb Z)
\cong H_2(A,\mathbb Z)\oplus H_2(B,\mathbb Z)
\oplus\bigl(H_1(A,\mathbb Z)\otimes H_1(B,\mathbb Z)\bigr).
$$
A <cyclic group> has zero second integral <group homology>, while
$$
C_m\otimes_{\mathbb Z}C_n\cong C_{\gcd(m,n)}.
$$
Consequently
$$
M(C_2\times C_4\times C_6)
\cong C_{\gcd(2,4)}\oplus C_{\gcd(2,6)}\oplus C_{\gcd(4,6)}
\cong C_2\oplus C_2\oplus C_2.
$$
Back to article page
