Order of the first cohomology group and subgroups

Question

Let $M$ be a $G$-module and $H$ a subgroup of $G$. Is $\# H^{1}(H, M) < \# H^{1}(G, M)$?

Answer 1

This is almost never the case. Here are two counterexamples:

1) Infinite groups: Let $G$ be a free group on $n$ generators, and let $H$ be a subgroup of $G$ isomorphic to a free group on $m$ generators, with $m\neq n$. Then one has $\mathrm{H}^1(G,\mathbb{Z})\cong \mathbb{Z}^n$ and $\mathrm{H}^1(H,\mathbb{Z})\cong \mathbb{Z}^m$.

2) finite groups: Let $G$ be the alternating group on $5$ elements and let $H$ be a subgroup of $G$ isomorphic to $\mathbb{Z}_2$. Then $\mathrm{H}^1(G,\mathbb{Z}_2)$ is zero since $G$ is perfect, but $\mathrm{H}^1(H,\mathbb{Z}_2)=\mathbb{Z}_2$.

Here are some cases in which it is true:

a) If $H$ has finite index in $G$, and $\mathrm{H}^{1}(G,M)$ is a finite group of order coprime with $[G:H]$, then one has (by transfer-restriction) an inclusion $\mathrm{H}^1(G,M) \rightarrow \mathrm{H}^1(H,M)$.

b) If $H$ is normal in $G$ then one has (by the LHS-spectral sequence) an exact sequence

$$0 \rightarrow \mathrm{H}^1(G/H,M^H)\rightarrow \mathrm{H}^1(G,M) \rightarrow \mathrm{H}^1(H,M)^G.$$ Hence, one obtains an inclusion $\mathrm{H}^1(G,M) \rightarrow \mathrm{H}^1(H,M)$ if and only if $\mathrm{H}^1(G/H,M^H)=0$. This is for example the case when $G/H$ has finite abelianization and $M^H$ is a torsion-free abelian group with trivial $G/H$-action, or when $M^H=0$.