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Show that $H \leq C_G(C_G(H))$.

We have that $C_G(H) = \left\{ g \in G | gh = hg \; \forall h \in H \right\}$. So then $C_G(C_G(H))$ is the $g$'s that commute with all $g$'s that commute with $h$. This would mean $$C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$$ But since $gg' \in G$ this is just some element of $G$, so doesn't this become the same as $C_G(H)$?

I feel like what I have is wrong but I am not sure why.

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Pick $h\in H$. We wish to show that $hg=gh$ for any $g\in C_G(H)$. But this is tautological from the definition of $C_G(H)$, so $H\subset C_G(C_G(H))$.

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  • $\begingroup$ How does $hg=gh \implies H\subset C_G(C_G(H))$? $\endgroup$ – user372834 Oct 20 '18 at 17:49
  • $\begingroup$ @user372834 Just by writing down the definition of the centralizer $\endgroup$ – TomGrubb Oct 20 '18 at 17:52
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in the definition $ C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$, is not $\forall g' \in G$, but $\forall g' \in C_G(H)$.

I believe that the question is wrong, because:

If $h_1\in C_G(C_G(H))$ and $h_1 \in H$, we have $$ h_1g'h_2=h_2g'h_1 ,\forall g' \in C_G(H), \forall h_2 \in H $$ It's clear that the identity $e \in C_G(H)$, so $$h_1h_2=h_2h_1, \forall h_2 \in H $$, then $H \subset C_G(C_G(H))$ only if H is abelian. But $H \leq C_G(C_G(H))$ also can indicate relation of order.

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  • $\begingroup$ The definition of the centralizer of the centralizer was indeed written incorrectly. But the statement to be shown was correct. And I have never seen anyone use $\leq$ for order between subgroups. $\endgroup$ – Tobias Kildetoft Oct 21 '18 at 7:22
  • $\begingroup$ I got it! C_G(C_G(H)) = { g \in G | gg' = g'g, forall g' in C_G(H)} and it's easy to show . Thank you! The answer above is correct and I do not saw it. $\endgroup$ – Max Will Oct 21 '18 at 14:30

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