Let $G$ be a group and $M,M',M''$ be $G$-modules. Currently, I am trying to understand the following section from Milne's Fields and Galois Theory (page 70):
Problem: I am trying to understand the maps of the second sequence, especially $d$.
In part, Milne tries to describe the map $d$. The part where I am stuck is why $\sigma m – m$ lie in the submodule $M'$ of $M$.
Let $f$ be the grpup homomorphism $M' \to M$ and $g$ be the group homomorphism $M \to M''$ from the first sequence. Also, let $m \in M$ such that $g(m) = m''$ (as Milne does, just without giving the map a name). We know that $M' \simeq \operatorname{im}(f) = \ker(g)$ because of the exactness of the first sequence.
So in order to show $\sigma m – m \in M'$ (for an arbitrary $\sigma \in G$), we must show $g(\sigma m – m) = 0$. I only got
$$g(\sigma m – m) = g(\sigma m) – g(m) = g(\sigma m ) – m''.$$
Here, I cannot find an argument why $g(\sigma m) = m''$. I also noted that I have not used the fact yet that $m'' \in M''^G$, i.e. $\tau(m'') = m''$ for all $\tau \in G$.
Could you please help me with that? Thank you!
Best Answer
You absolutely need the fact that $m''$ is fixed by $\sigma$. Thus in the quotient group, $\sigma m''-m''=0$. Thus, if $m$ is any preimage of $m''$ in $M$, although it need not lie in $M^G$, it does modulo $M'$. What this means is that $g(\sigma m-m)=\sigma(gm)-gm=\sigma m''-m''=0$. Thus $\sigma m-m\in \ker g=M'$, as needed.
I think the bit you missed is that $g$ is a homomorphism of $G$-modules, i.e., $g(\sigma m)=\sigma(gm)$.