# Which reflection groups can be enlarged?

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

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

## 1 Answer

$$A_1^*$$ 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. • I believe B_3 does have an outer automorphism, but not in$O(R^3)$. Feb 12, 2021 at 0:41 • @RichardLyons Good point, thanks. Feb 12, 2021 at 0:50 • @RichardLyons Can you say more about this? Feb 12, 2021 at 1:06 •$B_3=Z\times S$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)\$. Feb 12, 2021 at 2:15