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This is a result of Artin and Tate found in Serge Lang's Algebra. (Page 166.)

Let $G$ be a finite group operating on a set $S$. For $w\in S$, denote $1\cdot w$ by $[w]$, so that we have the direct sum $$ \mathbb{Z}\langle S\rangle=\sum_{w\in S}\mathbb{Z}[w]. $$ Define an action of $G$ on $\mathbb{Z}\langle S\rangle$ by defining $\sigma[w]=[\sigma w]$, and extending $\sigma$ to $\mathbb{Z}\langle S\rangle$ by linearity. Let $M$ be a subgroup of $\mathbb{Z}\langle S\rangle$ of rank $\#(S)$. Show that $M$ has a $\mathbb{Z}$-basis $\{y_w\}_{w\in S}$ such that $\sigma y_w=y_{\sigma w}$ for all $w\in S$.

I know this is an adaptation of a result in Artin and Tate's notes on Class Field Theory. I found proofs of similar ideas in Lang's Algebraic Number Theory (Theorem 1 of page 190), and in Nancy Childress's book here.

I tried reading through the proofs and the relevant lemmas, but have had difficulty deciphering them and putting them in the context of Lang's statement. Is there perhaps a dumbed down proof of the above statement more suited to using knowledge of basic groups, rings, and modules? Thanks.

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up vote 4 down vote accepted

I must be badly misunderstanding the statement, since the way I read it, it is completely wrong. It seems to be saying that every full rank subgroup of a permutation module over $\mathbb{Z}$ is itself isomorphic to a permutation module, which is not true.

Firstly, not every full rank subgroup is even a submodule in the first place. Examples abound: e.g. take the regular action of $G$ on itself (so $S=G$) and consider the submodule of $\mathbb{Z}[G]$ given by $\{\sum_{g\in G}a_g g: a_g\in \mathbb{Z}, a_1\in 2\mathbb{Z}\}$. This is a subgroup, but is certainly not preserved by the regular $G$-action.

Secondly, not every submodule is a permutation module. E.g. take $G=C_2$, the cyclic group of order 2, and consider the regular action. One can visualise the module $\mathbb{Z}[G]$ as a rectangular lattice on which the non-trivial element $g$ of $G$ acts by reflection in a diagonal (the obvious basis consists of $[1]$ and $[g]$). Now, consider the submodule spanned by $[1]+[g]$ and $[1]-[g]$. This is a direct sum of two one-dimensional modules over $\mathbb{Z}$, and is therefore not isomorphic to the original module, which is in contradiction to what the statement is claiming.

Edit: Ok, I had a look at the reference in Lang's Algebraic Number Theory, and there he does give a correct version of this statement: every $G$-stable full rank sublattice of a real permutation module has a full rank submodule isomorphic to a permutation module. But that's not at all what the present exercise is saying.

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I was having a hard time finding a sensible approach, so this is reassuring. Thanks for looking into it, Alex. – yunone Oct 21 '11 at 7:05
Dear yunone, you are welcome! – Alex B. Oct 21 '11 at 7:24

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