# Normalizer and Centralizer

• Oct 8th 2011, 04:18 PM
dwsmith
Normalizer and Centralizer
$H\leq G$ and $|H|=2$. Show that $N_G(H)=C_G(H)$. Deduce that if $N_G(H)=G$, then $H\leq Z(G)$

If $h\in H$, then $|h|=2$ if $h\neq e$.
$N_G(H)=\{g\in G:gHg^{-1}=H\}$
$C_G(H)=\{{g\in G:ghg^{-1}=h, \ \forall h\in H\}$.

How can I get to the point where they are equal with this information?
• Oct 8th 2011, 08:15 PM
Deveno
Re: Normalizer and Centralizer
well it's obvious that $C_G(H)$ is a subgroup of $N_G(H)$. so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

so let g be in $N_G(H)$. clearly $geg^{-1} = e$, and we know that $ghg^{-1} \in H$, where h is the non-identity element of H.

but if $ghg^{-1} = e$, then gh = g, and thus h = e, contradiction. hence $ghg^{-1} = h \implies gh = hg$, so $g \in C_G(H)$
• Oct 8th 2011, 08:15 PM
dwsmith
Re: Normalizer and Centralizer
Quote:

Originally Posted by Deveno
well it's obvious that $C_G(H)$ is a subgroup of $N_G(H)$. so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

so let g be in $N_G(H)$. clearly $geg^{-1} = e$, and we know that $ghg^{-1} \in H$, where h is the non-identity element of H.

but if $ghg^{-1} = e$, then gh = g, and thus h = e, contradiction. hence $ghg^{-1} = h \implies gh = hg$, so $g \in C_G(H)$

I am good thanks. I marked the thread solved awhile ago.