Normalizer and Centralizer

$\displaystyle H\leq G$ and $\displaystyle |H|=2$. Show that $\displaystyle N_G(H)=C_G(H)$. Deduce that if $\displaystyle N_G(H)=G$, then $\displaystyle H\leq Z(G)$

If $\displaystyle h\in H$, then $\displaystyle |h|=2$ if $\displaystyle h\neq e$.

$\displaystyle N_G(H)=\{g\in G:gHg^{-1}=H\}$

$\displaystyle C_G(H)=\{{g\in G:ghg^{-1}=h, \ \forall h\in H\}$.

How can I get to the point where they are equal with this information?

Re: Normalizer and Centralizer

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

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

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

Re: Normalizer and Centralizer

Quote:

Originally Posted by

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

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

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

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