Normalizer and Centralizer

**dwsmith** · October 8th 2011, 03:18 PM

$H\leq G$ 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?

**Deveno** · October 8th 2011, 07:15 PM

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 $g \in C_G(H)$

**dwsmith** · October 8th 2011, 07:15 PM

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.

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