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Let $G$ be a finite group and let $Rep(G)$ be its representation ring (as a group it is the free $\mathbb{Z}$-module on the irreducible complex reps). The collection of permutation representations $\mathbb{C}[\mathcal{O}]$ for $\mathcal{O}\cong G/H$ a $G$-orbit generate a $\mathbb{Z}$-subalgebra which we will denote by $Per(G) \subset Rep(G)$.

If $Per(G) = Rep(G)$ then it follows that all complex representations of $G$ are defined over $\mathbb{Z}$. To see this note that under our assumption for every representation $V$ there exists a representation $U$ defined over $\mathbb{Z}$ s.t. $U \oplus V = W$ is defined over $\mathbb{Z}$. We can now take $V_{\mathbb{Z}} = W_{\mathbb{Z}} / U_{\mathbb{Z}}$ as a $\mathbb{Z}$-form for $V$ (this is not really a precise proof, maybe a better arguemnt would be to explicitly write the projection operator which projectss onto the isotypic component of $V$ inside $\mathbb{Z}[G]$).

Is the converse true?

Question: Suppose every $\mathbb{C}$-representation of $G$ has a $\mathbb{Z}$-form, does it follow that $Per(G) = Rep(G)$? If not what's a counter example?

As an example when $G=S_n$ both of the properties are satisfied and this is in fact the only non-trivial example I know of.

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    $\begingroup$ Just to make sure, isn't "defined over $\mathbf{Z}$" the same as "defined over $\mathbf{Q}$", as one can always choose a $G$-stable lattice inside a $\mathbf{Q}G$-module? $\endgroup$
    – Wille Liou
    Jul 2 '19 at 8:24
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    $\begingroup$ @WilleLiou I believe this is true. You can just take the sum of all translates of a lattice to get a $G$-stable lattice (because it is projective and projectives are free over $\mathbb{Z}$). $\endgroup$ Jul 2 '19 at 8:32
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The answer is no. Counterexamples include the Weyl groups of types E6, E7, and E8. For a proof that the representations of these Weyl groups are indeed all realisable over $\mathbb{Q}$ (equivalently over $\mathbb{Z}$), see M. Benard, On the Schur Indices of Characters of the Exceptional Weyl Groups, Annals of Mathematics, Vol. 94, No. 1 (Jul., 1971), pp. 89-107 (MSN). For a proof of the statement that $Per(G)\neq Rep(G)$ for these groups, see D. Kletzing, Structure and Representations of Q-Groups, Lecture Notes in Mathematics 1084 (MSN).

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The answer is no. For a discussion and counter example, see (for example)
MR3451397 Bartel, Alex; Dokchitser, Tim Rational representations and permutation representations of finite groups. Math. Ann. 364 (2016), no. 1-2, 539–558. https://arxiv.org/abs/1405.6616

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  • $\begingroup$ Indeed, the dihedral group of order 8 gives already a counter example. $\endgroup$
    – Wille Liou
    Jul 2 '19 at 9:07
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    $\begingroup$ @WilleLiou Actually, that's not true. As mentioned in the introduction of the above paper, there is no counter example when $G$ is a $p$-group. $\endgroup$ Jul 2 '19 at 9:17
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    $\begingroup$ @HenriJohnston If I understand correctly the paper shows that it is possible for $Rep_{\mathbb{Q}}(G) / Perm(G)$ to be non-trivial in general (i.e. the quotient of the ring of virtual rational representations by $Perm(G)$). Though it doesn't assume (as in my question) that $Rep_{\mathbb{Q}}(G) = Rep_{\mathbb{C}}(G)$. I tried to look there whether there's a counter example to this as well but I didn't find it. No doubt its there and I just missed it... $\endgroup$ Jul 2 '19 at 9:57
  • $\begingroup$ @SaalHardali sorry, I missed the $Rep_{\mathbb{Q}}(G) = Rep_{\mathbb{C}}(G)$ requirement. Of course, Alex's answer above fully answers your question. $\endgroup$ Jul 2 '19 at 17:31
  • $\begingroup$ @HenriJohnston No worries. I'm happy that you posted this answer if only for this very nice and readable paper. I would accept both answers if I could. $\endgroup$ Jul 2 '19 at 17:33

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