# A finite group $G$ all of whose reps are defined over $\mathbb{Z}$ and yet $Rep(G)$ is not generated by permutation representations

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.

• 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? Jul 2, 2019 at 8:24
• @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 $\mathbb{Z}$). Jul 2, 2019 at 8:32

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).
• @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. Jul 2, 2019 at 9:17
• @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... Jul 2, 2019 at 9:57
• @SaalHardali sorry, I missed the $Rep_{\mathbb{Q}}(G) = Rep_{\mathbb{C}}(G)$ requirement. Of course, Alex's answer above fully answers your question. Jul 2, 2019 at 17:31