# Isomorphism between reductive groups implies isomorphic Levi subgroups of prescribed type?

I've come up with the following question.

Assume that $G$ and $G'$ are two isomorphic reductive algebraic groups over an algebraically closed field $k$. If $P$ (resp. $P'$) is the standard parabolic subgroup of $G$ of type $\Theta\subset Dyn(G)$ (resp. of $G'$ of type $\Theta\subset Dyn(G')$), are the respective Levi subgroups $L_{P}$ and $L_{P'}$ of $P$ and $P'$ isomorphic? I think it should be the case but i didn't figure out a proof.

edit : here is what i tried : if $f$ is the isomorphism between $G$ and $G$. The Levi subgroup associated to the standart parabolic subgroup of type $\Theta$ in $G$ is the centralizer of $C_{G'}(T_{\Theta})$ of $T_{\Theta}$, where $T_{\Theta}=\left(\bigcap_{\alpha\in \Theta} \ker(\alpha)\right)$ ($\alpha$ are roots in $G$). Since we have $f(C_G(T_{\Theta}))=C_{G'}(f(T_{\Theta}))$, it would suffice to show that $f(T_{\Theta})=\left(\bigcap_{\beta\in \Theta} \ker(\beta)\right)$ where the $\beta$ are the roots in $G$...

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As in @MattE's answer, this sounds tautological, even with your edit... is there some intent or issue that is not being expressed? (Sometimes one does not ask the question one truly has in mind.) – paul garrett Jan 2 '13 at 23:22
it is clear that it holds with the edit, but it's not tautological with it, there is something to show. This is why i would like to know if there is another way. The main issue is that in the context i'm working, it was not clear that the identification btween the two $\Theta$ is given by $f$... However it seems to be the only natural way. – Louis La Brocante Jan 3 '13 at 1:22

You are writing $\Theta$ to denote a subset of the roots of both $G$ and $G'$; how are you identifying these two subsets? If the identification is via the isomorphism $f$, then the answer to your question is tautologically yes. If you have some other identification in mind, you should explain what it is.
Yes, if $T$ is the maximal torus of $G$, then i take $f(T)$ as maximal torus in $G'$ and take the root system associated to $(G',f(T))$. In this situation is the answer to my question completely obvious from the begining, or is it a consequence of what i wrote in the "edit" ? – Louis La Brocante Jan 2 '13 at 18:50
@Rolando: Dear Rolando, If you are using $f$ to identify the two $\Theta$'s, than everything is tautological. (If we identify the groups $G$ and $G'$ via $f$, then their diagrams, roots, maximal tori, Levi's, everything, are the same.) As Paul Garrett suggests, if there is something more going on in the context where this question arose, you should edit your question to indicate it, or perhaps ask a new, more detailed question (with mutual links between that question and this one). Regards, – Matt E Jan 3 '13 at 1:55