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 $I_1\oplus B_d \subset B_{d+1}$.

  • 1
    $\begingroup$ $D_4$ has an exceptional triality symmetry, for what it's worth. $\endgroup$ Feb 11, 2021 at 14:42
  • 1
    $\begingroup$ @SamHopkins Might this lead to its inclusion in $F_4$? $\endgroup$
    – M. Winter
    Feb 11, 2021 at 14:43
  • 1
    $\begingroup$ @LSpice: Yes, the Coxeter diagram (unlike the Dynkin diagram) does not have arrows, so has an order two symmetry. $\endgroup$ Feb 11, 2021 at 14:52
  • 3
    $\begingroup$ 1. $A_4 ^*$, $A_7 ^*$, and $A_8 ^*$ can be enlarged into $H_4$, $E_7$, and $E_8$, respectively. $\endgroup$ Feb 11, 2021 at 15:06
  • 2
    $\begingroup$ @Sam Yes, $\mathrm{Weyl}(F_4) = \mathrm{Weyl}(D_4) \rtimes S_3$ $\endgroup$ Feb 11, 2021 at 15:59

1 Answer 1


$A_1^*$ 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.

  • 2
    $\begingroup$ I believe B_3 does have an outer automorphism, but not in $O(R^3)$. $\endgroup$ Feb 12, 2021 at 0:41
  • $\begingroup$ @RichardLyons Good point, thanks. $\endgroup$
    – Will Sawin
    Feb 12, 2021 at 0:50
  • $\begingroup$ @RichardLyons Can you say more about this? $\endgroup$
    – M. Winter
    Feb 12, 2021 at 1:06
  • 3
    $\begingroup$ $B_3=Z\times S$ 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 $GL_3(R)$. $\endgroup$ Feb 12, 2021 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.