= Solution
Put $Q=G/H$ and write $M^H$ for the <invariant submodule>. The <five-term exact sequence in group cohomology> associated with the <Lyndon–Hochschild–Serre spectral sequence> is
$$
0\longrightarrow H^1(Q,M^H)
\xrightarrow{\operatorname{inf}}H^1(G,M)
\xrightarrow{\operatorname{res}}H^1(H,M)^Q
\xrightarrow{d_2}H^2(Q,M^H)
\xrightarrow{\operatorname{inf}}H^2(G,M).
$$
The <inflation map in group cohomology> composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The <restriction map in group cohomology> restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a <one-cocycle> $f$,
$$
q\cdot[f]=\left[h\longmapsto g\cdot f(g^{-1}hg)\right].
$$
This is independent of the lift and of the representative at the level of <cohomology>. Finally, the <transgression in group cohomology> extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a <group coboundary>.
For the application, choose free generators $x_1,\ldots,x_n$ of $F$ and normal generators $r_1,\ldots,r_n$ of $R$. Since a <finite nonabelian simple group> $K$ is a <perfect group>, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to1$ with trivial coefficients contains
$$
0\longrightarrow\operatorname{Hom}(K,\mathbb Z)
\longrightarrow\operatorname{Hom}(F,\mathbb Z)
\xrightarrow{\operatorname{res}}\operatorname{Hom}(R,\mathbb Z)^K
\longrightarrow H^2(K,\mathbb Z)
\longrightarrow H^2(F,\mathbb Z).
$$
The first term is zero because $K$ is finite, and the last term is zero because a <free group> has cohomological dimension one. It remains to prove that restriction is surjective.
The <invariant submodule> of homomorphisms $R\to\mathbb Z$ is exactly
$$
\operatorname{Hom}(R/[F,R],\mathbb Z).
$$
The images of the relators $r_1,\ldots,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda(r_1),\ldots,\lambda(r_n))^T$. Let $A$ be the <relator exponent-sum matrix>. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a <unimodular matrix>. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi:F\to\mathbb Z$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi(r_j)=\lambda(r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives
$$
H^2(K,\mathbb Z)=0.
$$
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