The norm map in group cohomology is an isomorphism if $M$ is a projective $G$-module

Question

This is exercise III$.1.1(c)$ in Brown's "Cohomology of Groups."

Let $G$ be a finite group, $M$ a $G$-module, and $\overline{N}:M_G\to M^G$ the norm map defined by $[m]\to Nm$, where $N=\sum_{g\in G}g$. Show that $\overline{N}$ is an isomorphism if $M$ is a projective $\mathbb{Z}G$-module.

Recall that $M_G$ is the quotient of co-invariants and $M^G$ is the subgroup of invariants. Part $(a)$ of the same exercise asks us to show that $\ker\overline{N}$ and $\operatorname{coker}\overline{N}$ are annihilated by $|G|$. Using this, and the fact that $M_G$ is a free $\mathbb{Z}$-module (I think), it follows that $M_G$ has no element of finite order, and by part $(a)$, the kernel of $\overline{N}$ must be trivial.

I'd like to apply a similar argument to the cokernel, but the argument is more subtle because even though $M^G$ is a free $\mathbb{Z}$-module, it does not immediately follow that $M^G/NM$ has no elements of finite order (or at least I can't see the argument).

I've thought about this far too long, and welcome solutions and/or hints. Perhaps I'm not even on the right track.

Thanks.

Answer 1

This follows immediately from the following two observations:

1. $\overline{N}$ is an isomorphism when $M=\mathbb{Z}[G]$.
2. $\overline{N}$ is an isomorphism for $M=\oplus_\alpha M_\alpha$ if and only if it is an isomorphism for each $M_\alpha$.

Answer 2

First, remark that there is no reason for $M_G$ to be a free $\mathbf Z$-module. For instance, if $M$ is a finite abelian group with the trivial $G$-action, then $M_G=M$, which is certainly not $\mathbf Z$-free.

Suppose $M$ is a projective $\mathbf Z[G]$-module; let's show that $N$ is an isomorphism.

By definition of projective objects, $M$ sits in a split exact sequence of $\mathbf Z[G]$-modules

$$0 \to M \to F \to L \to 0$$

where $F$ is free, and $L$ is also projective.

The norm map sits in a diagram with exact rows

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \dots & \ra{ } &H_1(G,M) & \ra{ } & M_G & \ra{ } & F_G & \ra{ } & L_G & \ra{ } & 0 & & \\ & & \da{} & & \da{N} & & \da{N} & & \da{N} & & \da{} & & \\ & & 0 & \ra{ } & M^G & \ras{ } & F^G & \ras{ } & L^G & \ras{ } & H^1(G, L) & \ras{ } & \dots \\ \end{array}$$

The top row continues to the left, and the bottom continues to the right.

Now when $G$ is finite, $\mathbf Z[G]$ is an induced $G$-module, and therefore free $\mathbf Z[G]$-modules are acyclic for cohomology. This implies also that projective $\mathbf Z[G]$-modules are acyclic. Therefore $H^1(G,L)=0$.

Recall also that, by definition of group homology as left derived functors, projective objects are acyclic for homology. Therefore $H_1(G,M)=0$, so the diagram is

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \dots & \ra{ } &0 & \ra{ } & M_G & \ra{ } & F_G & \ra{ } & L_G & \ra{ } & 0 & & \\ & & \da{} & & \da{N} & & \da{N} & & \da{N} & & \da{} & & \\ & & 0 & \ra{ } & M^G & \ras{ } & F^G & \ras{ } & L^G & \ras{ } & 0 & \ras{ } & \dots \\ \end{array}$$

Now, it is easy to see that $N: F_G \to F^G$ is an isomorphism, and by the usual argument so is $N: M_G \to M^G$. (In fact, the diagram only gives injectivity of $M_G \to M^G$ and surjectivity of $L_G \to L^G$; but, switch the roles of $M$ and $L$!)