# Equivariant splitting of short exact sequences with a $\mathbb {Z}/2\mathbb{Z}$-action

Let $$G = \mathbb{Z}/2\mathbb{Z}$$, and consider a short exact sequence of \emph{free} $$\mathbb Z$$-modules of finite rank endowed with a $$G$$-action

$$0\to \mathbb Z^n \to \mathbb Z^m \to \mathbb Z^k\to 0,$$

and suppose that the maps are also $$G$$-equivariant. Clearly the sequence splits as sequence of $$\mathbb Z$$-modules, the question is:

does the SES split equivariantly?

Since, using the representation theory of $$G$$, any irreducible representation over $$\mathbb Z^n$$ has rank at most $$2$$, then I think that we can reduce to the case when $$k=2$$.

However I expect this to be well known to algebraists.

• By-the-way, is it true that any representation is a sum of one and two dimensional ones? Jul 3, 2023 at 17:59
• @Galoisgroup yes, see Charles W. Curtis and Irving Reiner. Representation theory of finite groups and associative algebras. Theorem 74.3 Jul 4, 2023 at 8:37

No. Consider $$0 \to \mathbb Z\to \mathbb Z\oplus \mathbb Z\to \mathbb Z\to 0$$

Here the action is trivial on the first $$\mathbb Z$$, permutation of coordinated on the second, and the multiplication on $$-1$$ on the third. The maps are $$f_1(a)=(a,a), f_2(a,b)=a-b$$ The reason is that the second $$\mathbb Z\oplus \mathbb Z$$ is not sum of two $$1$$-dimensianal representations.