= Solution
Let $P_\bullet\to\mathbb Z$ be a <projective resolution> of the trivial $\mathbb ZG$-module. The <projective-resolution definition of group cohomology> is
$$
H^n(G,M)=H^n\!\left(\operatorname{Hom}_{\mathbb ZG}(P_\bullet,M)\right).
$$
This is independent, up to a natural <isomorphism>, of the chosen <projective resolution>.
The degreewise natural isomorphisms
$$
\operatorname{Hom}_{\mathbb ZG}(P_j,M_1\oplus M_2)
\cong\operatorname{Hom}_{\mathbb ZG}(P_j,M_1)
\oplus\operatorname{Hom}_{\mathbb ZG}(P_j,M_2)
$$
commute with the <coboundary maps>. Taking <cohomology> proves that <group cohomology commutes with finite direct sums>:
$$
H^n(G,M_1\oplus M_2)
\cong H^n(G,M_1)\oplus H^n(G,M_2).
$$
Now restrict $P_\bullet$ from $G$ to a subgroup $K$. The <group ring> $\mathbb ZG$ is free as a $\mathbb ZK$-module, so restriction carries <free modules> to free modules and <projective modules> to projective modules. Thus the restricted complex is a <projective resolution> of the trivial $\mathbb ZK$-module. For the <coinduced module> $Y=\operatorname{Hom}_{\mathbb ZK}(\mathbb ZG,X)$, the <Hom functor adjunction for a coinduced module> gives an isomorphism of <cochain complexes>
$$
\operatorname{Hom}_{\mathbb ZG}(P_\bullet,Y)
\cong\operatorname{Hom}_{\mathbb ZK}(P_\bullet,X).
$$
Explicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb ZK$-linear map $a$ to $p\mapsto(r\mapsto a(rp))$. Taking <cohomology> proves <Shapiro's lemma>:
$$
H^n(G,Y)\cong H^n(K,X).
$$
For the <conjugation module of a group ring> $N=\mathbb ZG_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its <conjugacy classes>. Hence $N$ is the <direct sum> of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its <centralizer>. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb Z$. Applying <group cohomology commutes with finite direct sums> and <Shapiro's lemma> yields the <group cohomology of a conjugation module>:
$$
H^n(G,N)\cong\bigoplus_iH^n(C_G(g_i),\mathbb Z).
$$
The <symmetric group> $S_3$ has three <conjugacy classes>, represented by the identity, a transposition, and a three-cycle. Their <centralizers> are respectively
$$
S_3,\qquad C_2,\qquad C_3.
$$
For any <finite group> $L$ acting trivially on $\mathbb Z$,
$$
H^1(L,\mathbb Z)=\operatorname{Hom}(L,\mathbb Z)=0,
$$
because a <group homomorphism> sends an element of finite order to an element of finite order, while the <additive group> of the integers contains no nonzero <torsion elements>. Therefore
$$
H^1(S_3,N)=0.
$$
The <periodic resolution of a finite cyclic group> alternates the maps $t-1$ and $1+t+\cdots+t^{m-1}$. After applying $\operatorname{Hom}_{\mathbb ZC_m}(-,\mathbb Z)$ with the trivial action, these become alternately zero and multiplication by $m$, proving
$$
H^2(C_m,\mathbb Z)\cong\mathbb Z/m\mathbb Z.
$$
Combining this calculation with the supplied $H^2(S_3,\mathbb Z)\cong\mathbb Z/2\mathbb Z$ gives
$$
H^2(S_3,N)
\cong\mathbb Z/2\mathbb Z\oplus\mathbb Z/2\mathbb Z\oplus\mathbb Z/3\mathbb Z.
$$
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