I am not sure if this is the right place to ask but since I don't get many answers on this kind of questions on SE I might as well try it here. I also posted this question on SE.


I have the following problem and would be grateful if somebody could point me to a reference or give a brief explanation:



> 
> I have a semisimple algebraic group $G$ (you can assume in
>  characteristic 0 over an algebraically closed field or even
>  $\mathbb{C}$ if it helps) and an automorphism $\sigma:G\to G$ of order
>  2. Denote with $H$ the subgroup of $G$ fixed by $\sigma$ and let $N(H)$ be its normalizer in $G$. 
> 
> 
> The paper that I'm reading makes the following claim:
> 
> 
> The subgroup $K$ of $Aut(\mathfrak{g})$ generated by $Ad(N(H))$ and
>  $d\sigma$ is reductive.
> 
> 
> 


They assume that it's obvious that $Ad(N(H))$ is reductive and now since $Ad(N(H))$ has at most index 2 in $K$ we have that $K$ is reductive. Both statements are not quite clear to me right now and I couldn't find anything in the standard literature (I have Borels book and Tauvel/Yu's book on algebraic groups). 


Any pointers would be appreciated.



Automorphism of semisimple algebraic group of order 2
It is well-known that the fixed point group $H=G^\sigma$ for an involution is reductive and that it is of finite index in its normalizer. Search for "symmetric varieties". Therefore $K$ has a reductive subgroup of finite index and is therefore reductive itself.


The reductivity is best seen by observing that
$\mathfrak g=\mathfrak h\oplus\mathfrak p$ is a decomposition into eigenspaces of $\sigma$. Hence the restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ stays non-degenerate. This implies that $H$ is reductive.


