TITLE: Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group
QUESTION [7 upvotes]: I am looking for a reference for the following well-known fact: For any subdiagram $\Delta_0$ of of the Dynkin diagram $\Delta=D(G)$
of an adjoint simple group $G$ over an algebraically closed field $k$, there exists a reductive subgroup of maximal rank $G_0\subset G$
with Dynkin diagram $\Delta_0$.
To be more precise, I am looking for a reference for a proof of the following well-known lemma:
Lemma 1.  Let $G$ be an adjoint, connected, simple algebraic group with Dynkin diagram $\Delta=D(G)$
over an algebraically closed field $k$ of any characteristic.
Let $\Delta_0$ be a  subdiagram of $\Delta$
(that is, a subset $\Pi_0$ of the set $\Pi$ of vertices of $\Delta$,
together with all the edges of $\Delta$ connecting pairs of vertices of $\Pi_0$).
Then there exists a connected reductive $k$-subgroup of maximal rank  $G_0$ of  $G$
such that the corresponding adjoint semisimple group $G_0^{ad}$
has  Dynkin diagram $\Delta_0$.
I know a simple proof of Lemma 1, but I would prefer to give a reference rather than a proof.
The proof goes as follows. Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system,
then our $\Pi$ is a basis of $R$.  Let $S$ be the subgroup of $T$ orthogonal to $\Pi_0$,
then it is a subtorus of $T$ (because $G$ is adjoint). 
Set $G_0=C_G(S)$, the centralizer of $S$ in $G$.
Then $G_0$ is a connected reductive subgroup of $G$.
It is easy to see that (the adjoint group of) $G_0$ has Dynkin diagram $\Delta_0$.
Note that Lemma 1 is a special case of the following Lemma 2,
for which I would also be happy to have a reference.
Lemma 2.   Let $G$ be an adjoint, connected, simple algebraic group 
over an algebraically closed field $k$ of any characteristic.
Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system.
Let $R_0$ be a closed symmetric subset of $R$.
Then there exists a connected reductive $k$-subgroup of maximal rank  $G_0$ of $G$
with root system $R_0$.
I will be grateful to any references, comments, etc. (also to a proof of Lemma 2).
Mikhail Borovoi

REPLY [10 votes]: Dear Mikhail,
If I understand correctly, your Lemma 2 is implied by
SGA 3 Exposé 22, Théorème 5.4.7.
Everything is on a general base S (that you may take as your algebraically closed field). The kind of subgroup you want is called "de type (R)" (see Définition 5.2.1) and a subset of R that corresponds to such a group is also called "de type (R)". Now the theorem above exactly says that when a subset of R is closed, it is "de type (R)" which exactly means that there is a corresponding connected subgroup of G. By the way, Théorème 5.4.7 does not assume the subset to be symmetric, and you get things like Borel subgroups if you take only "half" of the roots. In the symmetric case, the group is reductive by Proposition 5.10.1.
Hope this helps.
Baptiste Calmès