Let $M,N$ be isomorphic as $\mathbf{Z}[G]$-mods, are they isomorphic as $\mathbf{Z}[H]$-mods, where $H<G$

Question

So I have recently been looking at the isomorphisms of $\mathbf{Z}[G]$-mods ($G$ finite), and noticed that a couple of my examples saw them isomorphic as $\mathbf{Z}[H]$-mods also, where $H$ is a subgroup of $G$. I was wondering if there is any general results for when this holds, such as when $H$ is a normal subgroup perhaps?

Answer 1

It holds without any condition on $H$, actually. Suppose $f : M \to N$ is an isomorphism of $\mathbb{Z}[G]$-modules. In other words, $f$ is bijective, additive and $G$-equivariant (ie. $f(g \cdot m) = g \cdot f(m)$ for all $g \in G$, $m \in M$). Then $f$ is also $H$-equivariant: $f(h \cdot m) = h \cdot f(m)$ for all $h \in H$, $m \in M$. It's still additive, so it's a morphism of $\mathbb{Z}[H]$-modules; and it's bijective, so it's an isomorphism.

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A more general fact holds. If $\phi : A \to B$ is a ring morphism, and $M$ is a $B$-module, let $\phi^* M$ be the same module viewed as an $A$-module through $\phi$. Then if $f : M \to N$ is an isomorphism of $B$-modules, the induced map $\phi^* M \to \phi^* N$ (on the level of sets it's the same map) is an isomorphism of $A$-modules, with the same proof as before. You can also view it as a purely formal result: $M \mapsto \phi^* M$ is a functor, and thus carries isomorphisms to isomorphisms.