Arbitrarily large finite irreducible matrix groups in odd dimension? I consider a finite irreducible matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^d)$. I am interested in the maximal size of $\Gamma$ depending on $d$. But this question makes only sense if there is an upper limit.
In even dimension there is no such limit. This is easiest seen in dimension $d=2$, where we have the cyclic groups or dihedral groups of arbitrarily large size. More generally, in dimension $d=2n$ we can consier the $n$-th cartesian power of a regular $k$-gon $P_k$:
$$\overbrace{P_k\times \cdots\times P_k}^{\text{n times}}.$$
Its symmetry group is irreducible and gets arbitrarily large with $k\to\infty$.

Question: What about odd dimensions? Can there be arbitrarily large finite irreducible matrix groups in dimension $d=2n+1$?

For example, in dimension $d=3$ we have the arbitrarily large symmetry groups of prisms and antiprisms, which are reducible. The largest irreducible group is probably the symmetry group of the icosahedron.
I have the feeling that in sufficiently large odd dimensions, the largest such group is the reflection group $B_d$.
 A: View your group as a subgroup $\Gamma$ of ${\rm GL}(d,\mathbb{C})$ by extending scalars. If it remains irreducible as such , then it has an Abelian normal subgroup $A$ with the index $[\Gamma :A]$ bounded in terms of $d$ (by a Theorem of Jordan). The rank of $A$ is clearly at most $d$, so we only need to consider the exponent of $A$.
Let $\chi$ be the character afforded by $\Gamma$. By Clifford's Theorem $\chi$ decomposes on restriction to $a$ as a sum $e(\lambda_{1} + \lambda_{2} + \ldots + \lambda_{t})$, where $et = d$ and each $\lambda_{i}$ is a linear character of $A$, with $\Gamma$ transitively permuting the $\lambda_{i}$.  Note that $e$ and $t$ are both odd. However, since $\chi$ is real-valued on restriction to $A$, we see that $\lambda_{i}$ and $\overline{\lambda_{i}}$  occur with equal multiplicity in ${\rm Res}^{\Gamma}_{A}(\chi)$ for each $i$. Since $t$ is odd, at least one $\lambda_{i}$ is real-valued.
But $[A:{\rm ker}\lambda_{i}]$ is independent of $i$ by the transitive action of $\Gamma$, and $A/{\rm ker} \lambda_{i}$  is cyclic for each $i$. Hence $[A: {\rm ker} \lambda_{i}] = 2$ for each $i$. Thus $A$ is an elementary Abelian $2$-group in the case that $\Gamma$ reamins irreducible as a complex linear group, and $|\Gamma|$ is bounded in terms of $d$ in that case. In this case, it is true for large enough $d$ that the maximum possible order is attained by the group $(\mathbb{Z}/2\mathbb{Z}) \wr S_{d}$, though this requires  theorems of B. Weisfeiler and M.J. Collins, which require the Classification of finite simple groups.
More generally, if the character $\chi$ afforded by $\Gamma$ is a sum of real-valued irreducible complex characters, we get a bound in terms of $d$ for $|\Gamma|$ by the same argument (applied to each irreducible summand).
Otherwise, (since we are given that $\Gamma$ is irreducible as a real linear group) we may write $\chi = \mu + \overline{\mu}$ for some non-real irreducible complex character $\mu$, contrary to the fact that $d$ is odd.
A: Indeed, in odd dimension it's bounded.
Indeed, let $\Gamma$ be such a matrix group. By Jordan's theorem, it has a normal abelian subgroup $\Lambda$ of index $\le c_d$. (An explicit bound for $d\ge 71$ is $c_d=(d+1)!$, by work of Collins and Weisfeiler, see Breuillard - An exposition of Jordan's original proof of his theorem on finite subgroups of $\operatorname{GL}_n(\mathbb C)$.)
If $\Lambda$ acts diagonalizably, then it has cardinal $\le 2^d$ and hence $\Gamma$ has cardinal $\le 2^dc_d$.
Otherwise, $\Lambda$ has blocks of size 2. By irreducibility, the sum of blocks of size two being invariant, it equals $\mathbf{R}^d$. Hence $d$ is even.
