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$C_G$ is the centralizer and $Z(G)$ is the center of $G$ if that wasn't clear. The only thing I can get from this is that $C_G(Z(G)) = Z(G)$ because if $gx=xg$ then $gag^{-1} = agg^{-1} = a$. I know that $Z(G) = G$ if $G$ is abelian, but $G$ isn't necessarily abelian in this case so that obviously doesn't help.

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3 Answers 3

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$$C_G(H)=\{g\in G:gh=hg\;\forall h\in H\}.$$

So what are the elements of $G$ which commute with every element of the centre? By definition, it's all of them.

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$C_G(H)$ = "The things in $G$ which commute with everything in $H$". So which things in $G$ commute with the center of $G$? By the definition of the center, it better be everything.

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Consider G to be a group.

By definition of center, Z(G) = {g∈G | gx = xg for all x∈G}

By the definition of centralizer , C_G(A) = { g∈G | gag^(-1) = a for all a∈A} Let g∈G, so by the definition of center, ga=ag for all a∈Z(G).

That is, for all a∈Z(G), we get gag^(-1).

This implies, g∈C_G(Z(G)), then we have G ⊂C_G(Z(G)) and C_G(Z(G))⊂ G.

We have proved that G=C_G(Z(G)).

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  • $\begingroup$ We have a Latex-like typesetting system for mathematical expressions, called MathJax. Information is here: math.meta.stackexchange.com/a/10164 $\endgroup$
    – 311411
    Commented Oct 30, 2023 at 17:53
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    – 311411
    Commented Oct 30, 2023 at 17:53

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