Is there any finite 8dimensional point group which contains the $E_8$ Coxeter group as a subgroup other than $E_8$ itself?

2$\begingroup$ What do you mean by symmetry group? Surely you don't want to include $O(8)$, the symmetry group of the sphere. Do you mean a finite subgroup of the orthogonal group? $\endgroup$ – Will Sawin Feb 11 at 3:29

$\begingroup$ Oops. I meant to add that but I must’ve forgotten. I’ve made an edit and fixed it. $\endgroup$ – Daniel Sebald Feb 11 at 3:51

$\begingroup$ Possibly related: mathoverflow.net/questions/37136/… $\endgroup$ – Sam Hopkins Feb 11 at 3:55

2$\begingroup$ What is a finite symmetry group? A finite reflection group? A finite subgroup of O(8)? $\endgroup$ – Moishe Kohan Feb 11 at 4:46

$\begingroup$ Please consider adding the tag "coxetergroups" to this question. $\endgroup$ – Nathan Reading Feb 11 at 15:42
No.
Let $G$ be a finite subgroup of $GL_n(\mathbb R)$ containing $W(E_8)$ as a subgroup. Because $G$ is compact, $G$ must preserve a symmetric positive definite form on $\mathbb R^8$. Since $W(E_8)$ preserves a unique such form, it must be that one.
Let $H$ be the largest subgroup of $G$ generated by reflections. Then $H$ contains $W(E_8)$ and thus is an irreducible reflection group, hence a Coxeter group. Examining the table of Coxeter groups and looking for entries in dimension $8$, there are four possibilities: $W(A_8), W(B_8), W(D_8), W(E_8)$. Because $W(E_8)$ has the highest order of these, we must have $H = W(E_8)$.
Now, by construction, $H$ is a normal subgroup of $G$, so $G$ normalizes $W(E_8)$, and hence $G$ is contained in the automorphism group of the $E_8$ root system. Because the CoxeterDynkin diagram $E_8$ has no nontrivial automorphisms, this is $W(E_8)$ and so $G= W(E_8)$, as desired.

$\begingroup$ In the last step, why should $G$ be a semidirect product? Even if it is, why should it inject into $\mathrm{Aut}E_8$? Also, $E_8$ does have a nontrivial outer automorphism (taking reflections to their negation). Diagram automorphisms induce group automorphisms, but are not always outer nor do they always exhaust the outer automorphisms. $\endgroup$ – Grant B. Feb 11 at 20:38

$\begingroup$ @GrantB. I said nothing about semidirect products. We can describe the $E_8$ root system as the set consisting of, for each reflection in $E_8$, the two vectors of length $2$ which are eigenvectors of the reflection with eigenvalue $1$. This makes it clear that any $G$ which normalizes $E_8$ stabilizes the $E_8$ root system, and hence, because there are no Dynkin diagram automorphisms, lies in $W(E_8)$. $\endgroup$ – Will Sawin Feb 11 at 21:58

$\begingroup$ @GrantB. Said to work in the generality of Coxeter groups, rather than just ADE type root systems: Consider the division of $\mathbb R^n$ by hyperplanes fixed by the reflections in a Coxeter group $H$. This splits $\mathbb R^n$ into a union of alcoves, on which $H$ and $N(H)$ (the normalizer of $H$ in $O(N)$, not its outer automorphism group) act. But $H$ acts transitively on alcoves because we can use the appropriate reflection to connect adjacent chambers, so every element of $N(H)$ is an element of $H$ times a stabilizer of your favorite alcove  i.e. an automorphism of the Coxeter diagram. $\endgroup$ – Will Sawin Feb 11 at 22:05

1$\begingroup$ @GrantB. Negation IS an element of the Coxeter group $E_8$. Specifically, it is the longest element. In general, negation is contained in a finite Coxeter group if and only if all of its exponents (see encyclopediaofmath.org/wiki/Coxeter_group ) are odd. $\endgroup$ – David E Speyer Feb 12 at 1:28

1$\begingroup$ @David Right, I agree that $1=w_0$ is in $E_8$, but the automorphism given by $s_i\mapsto s_i$ is still outer, since for instance the trace of each reflection changes sign. $\endgroup$ – Grant B. Feb 12 at 5:58