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.

  • $\begingroup$ $C_G(h)$ 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

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 $K=C_G(h)$.


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