# Order of the first cohomology group and subgroups

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

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What about $C_2\times C_2\lt A_4$, with $M$ the trivial $C_2$ module? – user641 Apr 16 '13 at 5:26
In general, if $M$ is a $p$-group (as abelian group) for a prime $p$, and $P \in {\rm Syl}_p(G)$ then, for any $n > 0$, $H^n(G,M)$ is isomorphic (using the restriction map) to a subgroup of $H^n(P,M)$, and it's often a proper subgroup. – Derek Holt Apr 16 '13 at 8:04
Wait, so is it in general true that $H^{1}(G, M)$ is a subgroup of $H^{1}(H, M)$? – ADF Apr 16 '13 at 11:42
No it is not true in general - it's easy to find counterexamples with $H=1$. – Derek Holt Apr 16 '13 at 13:49
What about $M=0$ ... – Martin Brandenburg Apr 17 '13 at 6:52

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 $M^H=0$.

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