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It is well-known that for a finite group $G$ and field $k$ of characteristic 0, the linearization morphism $B(G) \to R_k(G)$ has in most cases nontrivial kernel, and this can be used to find permutation modules which admit non-isomorphic permutation bases (as $G$-sets). I'm hoping to find a relatively easy-to-state example, but this time, for a permutation $\mathbb{Z} G$-module instead.

Unfortunately, all the examples I've cooked up over $\mathbb{Q}$ haven't extended to $\mathbb{Z}$. Given that Krull-Schmidt doesn't hold for $\mathbb{Z}G$ in general, I'm almost certain that examples of such permutation modules exist - does anyone have any examples of this phenomenon?

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    $\begingroup$ DB once told me that $Q_8$ ($Q_8$ quarternionic group of order 8) has $P$ (rank 8) such that $P \oplus P \cong \mathbb{Z}[Q_8]\oplus \mathbb{Z}[Q_8]$ for $P$ a non-trivial projective module. (Maybe knowing that is enough to reverse engineer $P$ -- or you could ping DB if you can reverse engineer the initials :) .) $\endgroup$
    – Geordie Williamson
    Commented May 16, 2023 at 5:35
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    $\begingroup$ I think that, in general, this occurs for all finite groups $G$ - the projective $\mathbb{Z}G$-modules have rank $|G|$, and for each projective, if it is summed with itself enough times, it becomes free! But I don't think I've ever actually seen an example of such a projective written down. I think all this should follow from Swan's theorem, but I could be wrong. $\endgroup$
    – Sam K
    Commented May 16, 2023 at 6:03
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    $\begingroup$ GW: Please see below. $\endgroup$
    – Dave Benson
    Commented May 16, 2023 at 11:27

2 Answers 2

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A similar question was asked on math.stackexchange a few years ago, and I posted the following answer. I've just looked again at Conlon's paper, and I'm afraid it's a bit short on explicit examples.

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There are fairly general examples due to Conlon in the paper:

Conlon, S.B., Monomial representations under integral similarity, J. Algebra 13, 496-508 (1969). ZBL0185.06702.

There is even a transitive example, due to Scott:

Scott, Leonard L., Integral equivalence of permutation representations, Sehgal, Surinder (ed.) et al., Group theory. Proceedings of the 21st biennial Ohio State-Denison mathematical conference, Granville, OH (USA), 14-16 May, 1992. Singapore: World Scientific. 262-274 (1993). ZBL0828.20004.

In this example, $G$ is $\text{PSL}(2,29)$ and the permutation actions are on the cosets of two non-conjugate subgroups both isomorphic to the alternating group $A_5$.

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    $\begingroup$ According to Remark 1 in Prasad, Dipendra, A refined notion of arithmetically equivalent number fields, and curves with isomorphic Jacobians. Adv. Math. 312, 198-208 (2017). Zbl 1430.11153, the above example of Scott is the `only' known example (i.e. ignoring examples that can be constructed from Scott's example in a trivial way). $\endgroup$
    – Henri Johnston
    Commented May 16, 2023 at 9:39
  • $\begingroup$ Thank you for the answer! I feel less bad about not being able to find any small examples with brute-force napkin math now :) $\endgroup$
    – Sam K
    Commented May 16, 2023 at 19:25
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This is too long for a comment, but isn't really an answer to the question. It's interesting anyway, so I'll leave this here.

For the record, the example Geordie mentioned is the Swan module $P$ for $\mathbb{Z}Q_8$ generated by $3$ and $N$ in the regular representation, where $N$ is the sum of the group elements. This is a non-free projective module of rank one, and it generates $\tilde K_0(\mathbb{Z}Q_8)\cong\mathbb{Z}/2$. We have $P\oplus P \cong \mathbb{Z}Q_8\oplus \mathbb{Z}Q_8$. As far as I can see, this doesn't lead to an example of the type requested, but it does lead to a permutation module with permutation bases that are inequivalent in an interesting way. Namely, if we let $M$ be a direct sum of four copies of $\mathbb{Z}Q_8$ then each of the three pairings of the four copies gives an isomorphism with a direct sum of four copies of $P$. Composing one of these with the inverse of another gives rise to inequivalent permutation bases of $M$.

But Jeremy's answer is better!

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    $\begingroup$ If you want an explicit projection $\mathbb{Z}Q_8\oplus\mathbb{Z}Q_8\to\mathbb{Z}Q_8\oplus\mathbb{Z}Q_8$ with image and kernel isomorphic to $P$, you can use the matrix $\left(\begin{smallmatrix}1-8N&21N\\-3N&8N\end{smallmatrix}\right)$. $\endgroup$
    – Dave Benson
    Commented May 16, 2023 at 9:16
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    $\begingroup$ This is quite interesting indeed - yet another way integral representations break all intuition for me. Thank you for the example! $\endgroup$
    – Sam K
    Commented May 16, 2023 at 19:23

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