Based on this question (which focuses on the case $E_8$) I wonder the following:

Question: For each finite reflection group $\Gamma\subseteq\mathrm O(\Bbb R^d)$, what is the largest finite group $\bar \Gamma\subseteq\mathrm O(\Bbb R^d)$ that contains $\Gamma$?

Some inclusions already happen among the reflection groups and their extensions:

  • For $I_2(n)$ there is no such largest group, because $$I_2(n)\subset I_2(2n)\subseteq I_2(4n)\subset\cdots\subset I_2(2^r n)\subset \cdots.$$
  • In general, we have $D_d\subset B_d$.
  • In general, $A_d\subset A_d^*$, where $A_d^*$ is the extended group that results from the additional symmetries of the Coxeter-Dynkin diagram of $A_d$.
  • We have $A_3=D_3\subset B_3=A_3^*$.
  • We have $A_4\subset A_4^*\subset H_4$.
  • We have $D_4\subset B_4\subset F_4\subset F_4^*$, again, $F_4^*$ is the extended group resulting from the additional symmetries of the Coxeter-Dynkin diagram.
  • We have $E_6\subset E_6^*$, for the same reason as above.
  • We have $A_7\subset A_7^*\subset E_7$.
  • We have $A_8\subset A_8^*\subset E_8$.

(Thanks to Daniel for noting the extensions of $A_4^*, A_7^*$ and $A_8^*$).

I believe that $H_3, H_4,B_5,E_7,E_8$ and $B_d,d\ge 9$ cannot be enlarged by the same reasoning as given in this answer (because these groups are the largest reflection groups in their respective dimension and their Coxeter-Dynkin diagrams have no additional symmetries).

So we are left with the following:

Question: Can we enlarge the groups $B_d$ ($d\in\{3,5,6,7\}$), $F_4^*$, $E_6^*$ and $A_d^*$ ($d\not\in\{4,7,8\}$)?

Maybe the inclusion $A_d\subset A_d^*\subset\cdots$ is not the right chain leading to the largest group. And I also have not touched on the reducible groups which can also be enlarged in some cases, e.g. the Coxeter-Dynkin diagram of $I_2(n)\oplus I_2(n)$ has also additional symmetries. We also have inclusions like $I_1\oplus B_d \subset B_{d+1}$.

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    $\begingroup$ $D_4$ has an exceptional triality symmetry, for what it's worth. $\endgroup$ – Sam Hopkins Feb 11 at 14:42
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    $\begingroup$ @SamHopkins Might this lead to its inclusion in $F_4$? $\endgroup$ – M. Winter Feb 11 at 14:43
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    $\begingroup$ @LSpice: Yes, the Coxeter diagram (unlike the Dynkin diagram) does not have arrows, so has an order two symmetry. $\endgroup$ – Sam Hopkins Feb 11 at 14:52
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    $\begingroup$ 1. $A_4 ^*$, $A_7 ^*$, and $A_8 ^*$ can be enlarged into $H_4$, $E_7$, and $E_8$, respectively. $\endgroup$ – Daniel Sebald Feb 11 at 15:06
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    $\begingroup$ @Sam Yes, $\mathrm{Weyl}(F_4) = \mathrm{Weyl}(D_4) \rtimes S_3$ $\endgroup$ – Theo Johnson-Freyd Feb 11 at 15:59

$A_1^*$ is maximal.

$A_2 = I_3$ and $A_2^* = I_6 = G_2$ (for the first one, consider the Dynkin diagram, for the second one, any group containing it as an index $2$ subgroup must normalize it, hence must be the normalizer) so $A_2^*$ is not maximal.

$A_3 = D_3$ and $A_3^* = B_3$. $B_3$ has order $48$, a multiple of $16$, while the only larger $3$-dimensional reflection group is $H_3$, of order $120$, not a multiple of $16$, so $B_3$ is a maximal reflection group (and thus maximal, because it has no outer automorphisms EDIT: in $O_3$).

But $A_n$ doesn't embed into $B_n$ for $n\geq 4$ because the alternating group on $n+1$ letters is fine simple in this range, thus has no nontrivial homomorphism onto the symmetric group on $n$ letters, hence no nontrivial automorphism into the extension of the symmetric group on $n$ letters by an abelian group. So $A_n^*$ is maximal unless $A_n^*$ is contained in an exceptional reflection group. In particular, $A_n^*$ is maximal for $n \geq 9$ or $n=5$.

Because $F_4$ has order $1152$, and $H_4$ has order $14400$, which is not a multiple of $1152$, $F_4$ does not embed into $H_4$. It also doesn't embed into a maximal $A_4$ or $B_4$ since those are smaller, so $F_4$ is a maximal reflection group, and thus $F_4^*$ is a maximal symmetry group.

I'm not sure why $B_5$ was on your list as there's no exceptional group in dimension $5$. Maybe you meant $B_8$.

$A_6$ can't embed into $E_6$ because the order of $E_6$ is not divisible by $7$. $B_6$ can't embed into $E_6$ because the order of $E_6$ is $72 \cdot 6!$ and thus is not divisible by $2^6 \cdot 6!$. $E_6$ can't embed into the other two since it has the largest order. So these are all maximal reflection groups and their normalizers are maximal symmetry groups.

The order of $E_7$ is $72 \cdot 8!$ which is divisible by the order of $B_7= 2^6 \cdot 7! = 2^3 \cdot 8!$, with quotient $9$. But the Weyl group of $E_7$, mod the center of order $2$, is simple, and can't have a subgroup of order $9$. if it did, its order would divide $9!\cdot 2$. So B_7$ is maximal.

According to Daniel Sebald, $B_8$ is not contained in $E_8$. Thus $B_8$ is maximal.

In summary, $A_n^*$ is maximal for all $n$ except $2$, $4$, $7$, $8$, $B_n$ is maximal for all $n$ except $2,4$, $I_n$ is never maximal, and the normalizer of every exceptional group is maximal.

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    $\begingroup$ I believe B_3 does have an outer automorphism, but not in $O(R^3)$. $\endgroup$ – Richard Lyons Feb 12 at 0:41
  • $\begingroup$ @RichardLyons Good point, thanks. $\endgroup$ – Will Sawin Feb 12 at 0:50
  • $\begingroup$ @RichardLyons Can you say more about this? $\endgroup$ – M. Winter Feb 12 at 1:06
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    $\begingroup$ $B_3=Z\times S$ where $|Z|=2$ (generated by the -1 mapping) and $S=D_3\cong S_4$, the symmetric group. There is a unique surjective homomorphism $\sigma:S\to Z$, the sign homomorphism. Then the mapping $\alpha:(z,s)\mapsto (z\sigma(s),s)$ is an automorphism of $Z\times S\cong B_3$. There are reflections in $S$, and they are carried by $\alpha$ to their negatives, so $\alpha$ is not inner -- it's not even conjugation by any element of $GL_3(R)$. $\endgroup$ – Richard Lyons Feb 12 at 2:15

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