The question is a very little more than what's in the title. It is easy (for some values of ‘easy’) to produce examples of endoscopic groups that are not subgroups. When I asked a colleague, he mentioned $\mathrm{PGL}\_3$ as an endoscopic group of $\mathrm G\_2$. However, in this example, $\mathrm{PGL}\_3$ is isogenous to $\mathrm{SL}\_3$, which is an actual subgroup of $\mathrm G\_2$, and similarly for the other examples that I know. Is this kind of ‘central failure’ as bad as it can get, or is there an example of an endoscopic group that is not even *isogenous* to a subgroup?



Endoscopic group that is not a subgroup
If I understand the definition correctly, a connected reductive group $H$ is an endoscopic group for a connected reductive group $G$ if its Langlands dual $H^\vee$ is a connected centralizer in $G^\vee$.


So $H=SO(2p+1)\times SO(2q+1)$ is endoscopic in $G=SO(2p+2q+1)$ since $H^\vee=Sp(2p)\times Sp(2q)$ is a centralizer in $G^\vee=Sp(2p+2q)$. But clearly $H$ is not isogenous to a subgroup of $G$. Another example is $H=SO(2p)\times Sp(2q)$ for $G=SO(2p+2q)$.


Using Borel-de Siebenthal the problem comes from the fact that for finite Dynkin diagrams one may have
$$
\text{extended dual}\ne\text{dual extended}.
$$


