3
$\begingroup$

Let $G$ be a finite group, $H<G$ be a subgroup, and let $h \in H$. Denote by $C_H(h)=\{x\in H: xh=hx\}$ and $C_G(h)=\{y\in G: yh=hy\}$ the centralizers of $h$ in groups $H$ and $G$, respectively.

Question: Is the following inequality always true: $$ \frac{|C_G(h)|}{|C_H(h)|} \le \frac{|G|}{|H|} $$ In our calculation we noticed that this holds for $G=S_n$, $H=S_k \times S_{n-k}$, so I was curious if this holds in general.

$\endgroup$
  • $\begingroup$ $C_G(h)$ is the stabilizer of the point $h$ in the action of $G$ on itself by conjugation and its index $|G/C_G(h)|$ is the length of the orbit containing $h$, which is the set of conjugates of $h$ in $G$. You get your inequality from the fact that every conjugate of $h$ under the smaller group $H$ is also a conjugate under $G$ (orbits don't become longer if you restrict the action to a smaller group). $\endgroup$ – j.p. Feb 14 '18 at 7:13
6
$\begingroup$

Yes. This is a special case of the following:

If $H$ and $K$ are subgroups of $G$, then $|G|\geq |HK|=|H||K|/|H\cap K|$, so $|G|/|H|\geq |K|/|H\cap K|$.

(Note that $HK$ might not be a subgroup of $G$, but the result still holds.)

In you case, you have $K=C_G(h)$.

$\endgroup$

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.