# Is the alternating group $A_8$ a subgroup of the exceptional lie group $G_2$?

Does $$G_2$$ have an $$A_8$$ subgroup? I think the answer is yes and that moreover this $$A_8$$ subgroup is maximal among the closed subgroups of $$G_2$$.

Some circumstantial evidence is that $$A_8$$ has a degree 7 irrep with Frobenius–Schur indicator 1 so it is a subgroup of $$SO_7$$ and $$G_2$$ is also a subgroup of $$SO_7$$. Also $$A_8$$ has a degree 14 irrep with Frobenius–Schur indicator 1, which could very possibly be the action of $$A_8$$ by the adjoint representation on the Lie algebra of $$G_2$$, which has dimension 14 ( a finite subgroup of a connected Lie group like $$G_2$$ being maximal among the closed subgroups is closely related to acting irreducibly in the adjoint representation).

This is all just conjecture though, I'm not familiar enough with $$G_2$$ to prove any of this. For what it's worth generators for the 7d irrep of $$A_8$$ are given

https://brauer.maths.qmul.ac.uk/Atlas/alt/A8/gap0/A8G1-Zr7B0.g

and generators for the 14d irrep of $$A_8$$ are given

https://brauer.maths.qmul.ac.uk/Atlas/alt/A8/gap0/A8G1-Zr14B0.g

Especially for the $$7 \times 7$$ generators I would imagine someone out there can "recognize" when matrices are " $$G_2$$ matrices" the same way that one can "check" if a matrix is symplectic.

$$G_2$$ is famously the group of symmetries of an antisymmetric trilinear form in its 7 dimensional representation.

The 7 dimensional representation $$V$$ of $$A_8$$ does not preserve any antisymmetric trilinear form. Preserving an antisymmetric trilinear form would correspond to an invariant vector in $$\Lambda^3(V)$$, but $$\Lambda^3(V)$$ is irreducible.

• Very interesting! Could you say more about why $\Lambda^3(V)$ is irreducible for $V$ the 7d irrep of $A_8$? Is it the 35d irrep of $A_8$? Oct 21, 2022 at 0:38
• Yes. If you take the standard $n-1$ dimensional representation $V$ of $S_n$ then all of its exterior powers $\Lambda^k(V)$ are irreducible, combinatorially these correspond to so-called hook shaped partitions $(n-k, 1^k)$. For $A_n$ the story is similar with the slight exception that if $n = 2k+1$ is odd then $\Lambda^k(V)$ decomposes into two parts.
– Nate
Oct 21, 2022 at 14:46
• +1 Wow that's such a cool fact I'll definitely remember that. I'd upvote it twice if I could! Oct 21, 2022 at 15:04
• can this fact be used to conclude that not only is $A_{n+1}$ a subgroup of $SO_n(\mathbb{R})$ but moreover the action of $A_{n+1}$ on $\mathfrak{so}_n(\mathbb{R})$ in the adjoint representation is irreducible for $n \neq 4$? (Since the adjoint rep here coincides with $\Lambda^2$ the second exterior power of the standard representation) Nov 10, 2022 at 22:50

No

There is no $$A_8$$ subgroup of $$G_2$$. According to 1.4 Corollary 2 of

Basic conjugacy theorems for $$G_2$$

All finite subgroups are either subgroups of positive dimensional subgroups like $$SU_3:2$$ or $$SU_2 \otimes SU_2 \cong SO_4$$ or they are from a short list of finite subgroups. These positive dimensional subgroups contain some interesting finite subgroups related to $$A_5,A_6,GL(3,2)$$ but nothing related to $$A_8$$. The short list of finite subgroups of $$G_2$$ is given in part (2) as $$GL(3,2),2^3\cdot GL(3,2),GL(3,2):2\cong SL(2,7), G_2(2)\cong Aut(PSU(3,3)),G_2(2)'\cong PSU(3,3), SL(2,8), PSL(2,13)$$ none of these are $$A_8$$. Indeed $$A_8$$ has order 20,160, and so it appears to be significatly larger than any irreducible finite subgroup of $$G_2$$ (the largest of which $$G_2(2)$$ has size 12,096). Thus there is no $$A_8$$ subgroup of $$G_2$$.