Does the (simply connected compact) Lie group $E\_7$ contain a finite subgroup $G \subset E\_7$ such that the $56$-dimensional irrep of $E\_7$ splits over $G$ as $28 \oplus \overline{28}$, but the $133$-dimensional adjoint representation remains simple when restricted to $G$? By "$28 \oplus \overline{28}$" I mean of course a $28$-dimensional complex irrep plus its dual.


Such a subgroup is definitely Lie primitive (meaning it doesn't fit inside a proper Lie subgroup $L \subset E\_7$), since an imprimitive subgroup would split $133 = \mathfrak{l} \oplus (\dots)$. According to [Griess and Ryba - Finite simple groups which projectively embed in an exceptional Lie group are classified!](http://www.ams.org/bull/1999-36-01/S0273-0979-99-00771-5), of the (quasi)simple subgroups of $E\_7$, only three are primitive, and none of them work. ($133$ splits over $SL\_2(29)$ and $SL\_2(37)$, and $56$ remains simple over $PSU\_3(8)$.) But I don't know about nonsimple subgroups.


This feels like a good homework problem, but in fact came up in my research: such a subgroup would allow me to build a superconformal field theory with some nice properties.



Can the defining rep of $E_7$ split over a finite subgroup while the adjoint remains simple?
I know this is an old question, but I can answer it in the negative. First, I have mostly completed a list of the Lie primitive subgroups of $E\_7(k)$ for all $k$, including $k=\mathbb C$, and there is no such example.


But even before that, a paper of Liebeck and Seitz, entitled 'Subgroups of exceptional algebraic groups which are irreducible on an adjoint or minimal module', appearing in J. Group Theory, shows that only simple subgroups act irreducibly on the adjoint.


In fact, for $p=5$ there are examples, namely the sporadic groups $M\_{22}$, $HS$ and $Ru$. But there is none over $\mathbb{C}$.


