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

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.

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 $$H\subset C_G(C_G(H))$$.

• How does $hg=gh \implies H\subset C_G(C_G(H))$? – user372834 Oct 20 '18 at 17:49
• @user372834 Just by writing down the definition of the centralizer – TomGrubb Oct 20 '18 at 17:52

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.

• 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. – Tobias Kildetoft Oct 21 '18 at 7:22
• 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. – Max Will Oct 21 '18 at 14:30