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$G$ is a group and $H$ is a subgroup of $G$. Prove that $C_G(H)<N_G(H)$.

Missed the discussion on normalizer and centralizer. I just have the definitions: $$N_G(H)=\{g\in G\mid gHg^{-1}=H\}$$ $$C_G(H)=\{g\in G\mid gh=hg, \forall h\in H\}$$

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If $x\in C_G(H)$ then $xH=Hx$.

So $xHx^{-1}=H$.

Hence $x\in N_G(H)$.

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  • $\begingroup$ Well, that is simple. I think I was looking way too into this question. $\endgroup$ – maidel b Nov 26 '15 at 0:04
  • $\begingroup$ no problem my friend $\endgroup$ – janmarqz Nov 26 '15 at 0:05

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