Let $G$ be a group.

It is written in my text that there is a homomorphism $\phi:G\rightarrow Aut(H)$ where $H$ is a normal subgroup of $G$ and $\ker(\phi)=C_G(H)$.

From this, i realize that $C_G(H)$ is normal when $H$ is normal.

Is it a trivial result?

That is, can one deduce this from the definition with basic set operations? That is, showing $gC_G(H)g^{-1}\subset C_G(H)$?

What about $N_G(H)$? Is this normal?

  • $\begingroup$ $C_G(H)$ denotes the centralizer of $H$. $\endgroup$ – user156562 Jun 14 '14 at 18:33
  • 1
    $\begingroup$ $C_G(H)$ is normal because it is the kernel of a homomorphism of groups (easy exercise). What do you mean by $N_G(H)$? $\endgroup$ – Pgatti Jun 14 '14 at 18:43
  • $\begingroup$ Well @Pgatti, $\;C_G(H)\;$ is certainly normal... in $\;N_G(H)\;$ , not necessarily in $\;G\;$ ... $\endgroup$ – DonAntonio Jun 14 '14 at 18:51


$$\phi:G\to \text{Aut}\,H\;,\;\;\phi(g):=\Phi_g$$

where $\;\Phi_g\;$ is the inner automorphism of $\;H\;$ determined by $\;g\;$ , i.e.

$$\Phi_g(h):=g^{-1}hg\;,\;\;\forall\,h\in H$$

Since $\;H\lhd G\;$ the above is well defined, and we have that

$$\ker\phi:=\{g\in G\;:\;\Phi_g=Id_H\}=\{g\in G\;:\;g^{-1}hg=h\iff gh=hg\;\forall\,h\in H\}=:C_G(H)\;$$


In general, $C_G(H)$ is normal subgroup of $N_G(H)$ and when $H$ is normal,

$N_G(H)=G$ then the result follows.


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