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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.

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

1 Answer 1

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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.

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