0
$\begingroup$

I know this may be a basic question, but since Algebra is not my area, create examples is always a big issue for me.

I need an example of group $G$ and a subset $H$ of $G$ such that the centralizer of $H$, $C_G(H)=\{x \in G; xh=hx \ \forall \ h \in H\}$ is a strictly subgroup of the normalizer of $H$, $N_G(H)=\{x \in G;xH=Hx\}$ and it is, at the same time, a strictly subgroup of $G$.

I was trying to work with the Dihedral groups and the symmetric groups but I could not get anywhere.

If you know the example, you don't need to show me the justification, just give me a hint with the group and $H$ and I am gonna try justify by myself.

Thanks everyone.

$\endgroup$
1
$\begingroup$

Let $H$ be any non-abelian non-normal subgroup of $G$. Then:

  • $H$ is not contained $C_G(H)$. (If $H \subseteq C_G(H)$ then $H$ is abelian.) Therefore, $C_G(H) \ne N_G(H)$ because $N_G(H)$ always contains $H$.
  • $N_G(H) \ne G$ because $H$ is not normal.

As a concrete example, we may take the subgroup $S_3$ of $S_4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.