$\operatorname{Hom}_\Lambda (B, \operatorname{Hom}_{\Bbb Z}(\Lambda, X)) = \operatorname{Hom}_\Bbb Z (B, X)$ Let $G$ be a group, $B$ a $G$-module and $X$ an abelian group. Let $\Lambda:=\Bbb Z[G]$.
Serre states in his book local fields that we have the equality:
$$\operatorname{Hom}_\Lambda (B, \operatorname{Hom}_{\Bbb Z}(\Lambda, X)) = \operatorname{Hom}_\Bbb Z (B, X)$$
I know this must be easy, but I'm not sure if I'm seeing it for the right reason. Let $\phi\in \operatorname{Hom}_{\Bbb Z}(\Lambda, X), \lambda \in \Lambda$. To begin with, what is the $\Lambda$-module structure on $\operatorname{Hom}_{\Bbb Z}(\Lambda, X)$? Is it
$$\lambda\phi (\cdot)\mapsto \phi((\lambda)\,\cdot\,)?$$
Let $\Phi \in\operatorname{Hom}_\Lambda (B, \operatorname{Hom}_{\Bbb Z}(\Lambda, X))$. If the above is correct, then I get
$$\Phi (b) (\phi) (\lambda)=\Phi(b)(\lambda\phi)(1)=\lambda\Phi(b)(\phi)(1)$$
where the second equality follows from the fact that $\Phi$ is a $\Lambda$-hom. And I am stuck here.
Anyways, $\Bbb Z$-maps from $\Lambda$ to $X$ are determined by where you send each of the elements $G\ni g \mapsto x \in X$. So we have
$$\operatorname{Hom}_\Lambda (B, \operatorname{Hom}_{\Bbb Z}(\Lambda, X)) = \operatorname{Hom}_\Lambda (B, \operatorname{Hom}_{\Bbb Z}(G, X)) $$
 A: This follows from very general principles. If you have rings $R,S$, an $(R;S)$-bimodule $A$, a  left $S$-module $B$, and a left $R$-module $C$, then $A\otimes_{S}B$ is naturally a left $R$-module, $\mathrm{Hom}_{R}(A,C)$ a left $S$-module, and the canonical bijection
$$\mathrm{Hom}_{R}(A\otimes_S B,C)\xrightarrow{\,\sim\,} \mathrm{Hom}_{S}(B,\mathrm{Hom}_{R}(A,C))$$
is an isomorphism of abelian groups. The left $R$-module structure on $A\otimes_{S}B$ is, naturally enough, given by $r\cdot (a\otimes b)=(ra)\otimes b$, while the left $S$-module structure on $\mathrm{Hom}_{R}(A,C)$ is given by $s\cdot\phi=\psi$ where $\psi(a)=\phi(as)$. Of course, if there are additional compatible module structures on$A,B,C$, then these descend to the tensor product and the hom-sets, and the isomorphism above is one of multimodules (instead of just abelian groups). A reference for all this is Algebra I by N. Bourbaki.

In any case, let $B$ be a left $\Lambda$-module. Set $A=\Lambda$ which is a $(\Bbb Z;\Lambda)$-bimodule and $C=X$ which is a left $\Bbb Z$-module. Then the above is an isomorphism
$$\mathrm{Hom}_{\Bbb Z}(\underbrace{\Lambda\otimes_{\Lambda} B}_{\simeq B},X)\xrightarrow{\,\sim\,} \mathrm{Hom}_{\Lambda}(B,\mathrm{Hom}_{\Bbb Z}(\Lambda,X))$$
