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Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-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 $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, 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:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,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. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

For $a,b\in I$, the square-zero ideal condition givesThus the square-zero unit subgroup $1+I$ is abelian, andis a group isomorphism from the additive group of $I$.

Use the specified ring isomorphism $R/I\cong \mathbb{Z}G$. For $x\in \mathbb{Z}G$, choose a lift $r\in R$ and define $x\cdot a=ra$ for $a\in I$. Two lifts differ by an element of $I$, whose product with $a$ vanishes, so this is well defined. Right multiplication is handled identically. The two actions commute by associativity, making $I$ a bimodule. If $u\in R$ lifts $g\in G$, then $u$ is a unit: a lift $v$ of $g^{-1}$ makes both $uv$ and $vu$ elements of $1+I$, hence units, and a ring element with both a left and a right inverse is invertible. Conjugation therefore definesChanging $u$ by an element of $I$ does not change this expression because $I^2=0$. Moreover,so $a\mapsto 1+a$ is an isomorphism of $\mathbb{Z}G$-modules for these conjugation actions.

Let $R^{\times }\to (R/I{)}^{\times }$ be reduction on unit groups, and define $U$ as the inverse image of the distinguished subgroup $G\subseteq (\mathbb{Z}G{)}^{\times }$. Every $g\in G$ has a unit lift by the preceding argument, and the kernel consists exactly of the units congruent to $1$, namely $1+I$. Multiplication in $R$ therefore gives the group extensionChoose a set-theoretic section $s:G\to U$ with $s(1)=1$. Its extension cocyclesatisfies the two-cocycle identity by associativity. A different section changes $c$ by a group coboundary, so second group cohomology classifies group extensions gives a well-defined classThe same construction for $(R_1,I_1)$ gives $U_1$, an extension cocycle $c_1$, and $x_1=[c_1]\in H^2(G,1+I_1)$.

The answer to the final question is no. An abstract ring isomorphism $R_1\cong R$ need not carry the distinguished ideal $I_1$ to $I$, need not induce the identity under the two chosen identifications of the quotient rings with $\mathbb{Z}G$, and need not induce the prescribed $\mathbb{Z}G$-module isomorphism $\theta$. Hence it need not give an isomorphism of the two displayed group extensions, so it imposes no equality $\varphi (x_1)=x$. That equality does hold if the ring isomorphism has all three compatibility properties, because it then carries one extension cocycle to the other up to a group coboundary.

If $I^2=0$, then $1+I=\{1+a:a\in I\}$ is an abelian subgroup of the unit group. Multiplication satisfies $(1+a)(1+b)=1+a+b$, and $(1+a{)}^{-1}=1-a$, so $a\mapsto 1+a$ identifies the additive group of $I$ with $1+I$.