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In one literature, I encountered the following problem.

Suppose $G$ is a group and $H$ is its subgroup. $N_G(H)$ is the normalizer of $H$ in $G$. A number $c(h)$ is defined for every $h$ in $H$ this way:

  1. we first decompose $G$ using coset of $N_G(H)$: $G=\bigcup_{\alpha}s_\alpha N_G(H)$,
  2. $c(h)$ is defined to be $c(h)=\#\{s_\alpha,s_\alpha^{-1}hs_\alpha\in H\}$, where $\#$ means number of elements in a set.

Then, it is claimed that $$c(h)=\frac{|G|}{\#[h]_G}/\frac{|N_{G}(H)|}{\#[h]_{N_G(H)}}=\frac{|C_G(h)|}{|C_{N_G(H)}(h)|},$$ where $C_G(h)$ ($C_{N_G(H)}(h)$) is centralizer of $h$ in $G$ ($N_G(H)$) and $[h]_G$ ($[h]_{N_G(H)}$) is conjugate class in $G$ ($N_G(H)$) containing $h$.

I cannot prove that these two numbers are equal and cannot find a counter example. Can anyone help me on this?

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  • $\begingroup$ I dont think this is true. Take H as a Sylow 2 subgroup of $S_4$, say generated by (24) and (1234). Then H is self normalizing. Let h be the central element of H, namely (13)(24). Then $C_G(h)$ and $C_H(h)$ are both H, so your RHS is 1. But there is an element outside H that keeps h in H, specifically (234). So $c(H)>1$. $\endgroup$ – Steve D Dec 22 '16 at 9:22
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I don't think it is true. Let $G = S_4$, $h=(1,2)(3,4)$, and $H=C_G(h)$. So $|H|=8$.

Then $g^{-1}hg \in H$ for all $g \in G$ and $N_G(H)=H$, so $c(h)=3$. But $|C_G(h)|=|C_{N_G(H)}(h)|=8$.

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  • $\begingroup$ If $g^{-1}hg \in H$ for all $g \in G$, $N_G(H)=G$ $\endgroup$ – Chong Wang Dec 22 '16 at 8:59
  • $\begingroup$ The subgroup $\{ 1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3) \}$ is normal in $G$ and contains $h$, so $g^{-1}hg \in H$ for all $g \in G$. But $H$ is not normal in $G$, because $G$ has $3$ Sylow $2$-subgroups all conjugate to $H$. $\endgroup$ – Derek Holt Dec 22 '16 at 9:06

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