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.
 A: $G_2$ 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.
A: No
There is no $ A_8 $ subgroup of $ G_2 $. According to 1.4 Corollary 2 of
Basic conjugacy theorems for $ G_2 $
by Greiss https://link.springer.com/content/pdf/10.1007/BF01884298.pdf
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 $.
