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There is an exercise on page 142 of Humphreys' Linear Algebraic Groups:

Ex.1 Let $G$ be a connected algebraic group, $x \in G$ is semisimple. Must $C_G(x)$ be connected?

When $G$ is solvable, I think of another fact whose correctedness is proved on the book:

Let $H$ be a subgroup (not necessarily closed) of a connected solvable group $G$, $H$ consisting of semisimple elements. Then $C_G(H) = N_G(H)$ is connected.

So, suppose that $G$ is solvable, set $H = \langle x \rangle$. It appears that $C_G(x)$ is connected.

I think the general case could be reduced to the solvable case if for any $y \in C_G(x)$, I can find a Borel subgroup of $G$ containing both $x$ and $y$. (Then $y$ must be in $C_B(x)$ which is connected.)

I think the Borel subgroup could be found, may be through the method of Borel variety. But I have difficulty in this.

Another exercise on the same page is:

Ex.2 Let $G$ be a connected algebraic group. If $x \in G$ has semisimple part $x_s$, then $x$ is contained in the identity component of $C_G(x_s)$.

If the answer to Ex.1 is affirmative, then Ex.2 would be obvious, since $x \in C_G(x_s)$. But I am not sure.

So, would you please tell me the answer to Ex.1, or help me with the proof or counterexample? If the centralizer could be not connected, how can I give Ex.2 a proof?

Sincere thanks.

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Dear ShinyaSakai, As a hint for Ex. 2, did you consider the case when $G = GL_n$? Here the Jordan decomposition is very explicit, and you can see directly how to answer Ex. 2 (without having to think at all about Ex. 1). Regards, –  Matt E Feb 13 '12 at 18:49
Also, regading Ex. 1, did you try any examples yourself? It is always good to think about $SL_n$ and $GL_n$ first. Also, since $SL_n$ is simply connected, it can be a little bit misleading sometimes, so it is good to think about $PGL_n$ too. Regards, –  Matt E Feb 13 '12 at 18:51
Dear @Matt E: Thank you for the comments. Ex.2 needs a proof, so do you mean the general case is done if it is proved in $GL_n$? For, Ex.2, I've thought of $GL_n$, in which semisimple elements can be considered as diagonal matrices. But it appears to me that the centralizer of a diagonal matrix in $GL_n$ is always connected. (For example, in $GL_3$, when $x$ is diagonal, with $1,2,1$ on the diagonal, has its centralizer isomorphic to $D_3 \times \mathbb{G}_a \times \mathbb{G}_a$, which is connected.) Is there any counterexample in $PGL_n$? Best regards, –  ShinyaSakai Feb 13 '12 at 19:36
Dear ShinyaSakai, Yes, it is true that centralizers of semisimple elements in $GL_n$ are connected, but why don't you try to prove Ex. 2 for $GL_n$ without using this fact, looking more directly at the Jordan decomposition. The point is that then you may find an approach which you can generalize to other groups. Regarding $PGL_n$, did you try looking for counterexamples? Why don't you try $n = 2$ first, and think concretely about what it means to centralize an element in $PGL_2$, by working upstairs in $GL_2$? Regards, –  Matt E Feb 14 '12 at 3:32

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