# Interpretation of first cohomology group

I was reading through Ivanov's book about 'The Monster group and Majorana involutions' when I bumped into the following statements on page 6.

Let $K$ be a group and $U$ be a $GF(2)$-module for $K$. If $W$ is the largest indecomposable extension of $U$ by a trivial module, then $W/U \cong H^1(K,U)$ and all the complements to $W$ in the semidirect product $W:K$ are conjugate and $H^1(K,W)$ is trivial. Dually, if $V$ is the largest indecomposable extension of a trivial module $V_0$ by $U$, then $V_0^\ast \cong H^1(K,U^\ast)$ (here $U^\ast$ is the dual module of $U$).

Can someone explain why these two statements hold? I have seen the definition of the cohomology groups in terms of the $Ext$-functor and I know the interpretation which uses derivations and principal derivations.

My first approach towards a proof is the following. Start with such a largest (not really sure what 'largest' should mean) indecomposable extension of $U$ by a trivial module $A$: $$0 \longrightarrow U \longrightarrow W \longrightarrow A \longrightarrow 0.$$ Turning this short exact sequence into a long one, we obtain: $$0 \longrightarrow H^0(K,U) \longrightarrow H^0(K,W) \longrightarrow H^0(K,A) \longrightarrow H^1(K,U) \longrightarrow H^1(K,W) \longrightarrow \dots.$$ Since $A$ is a trivial module, $H^0(K,A) \cong A$. If we proof that $H^1(K,W) = 0$ (which we need to proof anyway) and that $H^0(K,W)=0$, then we have indeed that $H^1(K,U) \cong A \cong W/U$. But $H^0(K,W)=0$ would imply that $H^0(K,U)=0$ which is definitely not the case for an arbitrary module $U$. So this suggests that this approach isn't exactly the right one or that some requirements on $U$ are missing.