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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, 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,2012-02-14

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