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.


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 $S_4$.


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