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Let, $G(k^{al})$ be an algebraic group, over an algebraically closed field, and $\Gamma_{G}$ is the set of all closed subgroups of $G(k^{al})$.

Then is the map $Z_{G}: \Gamma_{G} \rightarrow \Gamma_{G}$ which takes a closed subgroup to its centralizer in $G$, an involution? (probably not true)

If we now assume that $G(k^{al})$ is reductive or semisimple is there a characterization of all such closed subgroups for which $Z_{G}$ is an involution?

More generally if $G_{k}$ is an algebraic group scheme (now $k$ is no longer algebraically closed ) and $\Gamma_{G}$ is the set of closed group subschemes of $G_{k}$, do the previous two questions have a meaningful answer?

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up vote 3 down vote accepted

When $G=\mathrm{GL}_n$, then the centraliser $C$ of a subgroup scheme $H$ form the invertible elements of the algebra $M$ of matrices commuting with $H$. The group of such elements is Zariski dense in $M$ so $C$ and $M$ determine each other. Hence, the image of $Z_{\mathrm{GL}_n}$ and the question of involutivity on that image is completely reduced to double centraliser results for algebras.

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A comment on Torsten's answer: For any group $G$, and any subset $S \subset G$ it is true that $Z_G(Z_G(Z_G(S))) = Z_G(S)$. In particular the operation $Z_G$ is always an involution on its image. – Lucas Culler Feb 9 '14 at 16:22

Not necessarily. For cases when this does happen see the question 28354

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