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I have read on James E. Humphreys' Linear Algberaic Groups

If $G$ is an algebraic subgroup contained in $GL(n,K)$, and $s$ is a semisimple element of $G$, then $\mathfrak{g}$ has the decomposition: $\mathfrak{g} = \mathfrak{c}_{\mathfrak{g}}(s) \oplus \mathfrak{n}$, where $\mathfrak{c}_{\mathfrak{g}}(s) = \{ \mathrm{x} \in \mathfrak{g} | \mathrm{Ad}s \mathrm{x} = \mathrm{x}$ }. Moreover, $\mathfrak{c}_{\mathfrak{g}}(s)$ is the Lie algebra of $C_G(s)$, the centralizer of $s$ in $G$, and $\mathfrak{n}$ is the Lie algebra of $Cl_G(s)s^{-1}$, the elements of $G$ of the form $gsg^{-1}s^{-1},g\in G$.

Then I begin to wonder whether the product map $C_G(s) \times Cl_G(s)s^{-1} \rightarrow G$ is surjective.

For common abstract groups, set $G = S_3$, and $s = (123)$, and I see that $C_G(s) \times Cl_G(s)s^{-1} = \{ 1, (123), (132) \} \subsetneq G$. So the product morphism is not surjective.

For algebraic groups, let $G = SL(2, \mathbb{C})$, $s = \mathrm{diag}(2,1/2)$, then $C_G(s) = \{ \mathrm{diag}(d,d^{-1}) | d \in \mathbb{C}^* \}$, and $Cl_G(s)s^{-1} = \left\{ \begin{pmatrix} 1+\frac{3}{4}a_{12}a_{21} & -3a_{11}a_{12} \\ \frac{3}{4}a_{21}a_{22} & 1-3a_{12}a_{21} \end{pmatrix} | a_{11},a_{12},a_{21},a_{22} \in \mathbb{C}, a_{11}a_{22}-a_{12}a_{21} =1 \right\}$. After a lot of calculation, I find that $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \notin C_G(s) \times Cl_G(s)s^{-1}$ is probably true.

But, in the same book, I see

Let $\tau: C_G(s) \times Cl_G(s)s^{-1} \rightarrow G$ be the product morphism. Then $d \tau_{(e,e)}$ is an isomorphism of $\mathfrak{c}_{\mathfrak{g}}(s) \oplus \mathfrak{n}$ and $\mathfrak{g}$.

So, I want to know if $\tau$ is surjective for any algebraic group, as I don't have full confidence in my long calculation. Moreover, can we conclude that $C_G(s) \times Cl_G(s)s^{-1}$ has the same Lie algebra with $G$ from the surjectivity of $d \tau_{(e,e)}$? Then is it true that $C_G(s) \times Cl_G(s)s^{-1}$ contains the identity component of $G$? If this is true, there must be something wrong with my calculation, because $SL(2, \mathbb{C})$ is connected.

Sorry for so many questions, and I am thankful for any answer to, or hints for any one of them.

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