2
$\begingroup$

Let $G$ be a finite group, $N$ a subgroup of index $2$ of $G$. Consider the action of $G$ on itself by conjugation and the restriction of this action to $N$ (i.e. $N$ acts on $G$ by conjugation).

Let $x \in G$. We denote by $x^G = \{gxg^{-1} \mid g \in G\}$ the orbit of $x$ by the action of $G$ and $x^N = \{nxn^{-1} \mid n \in N\}$ the orbit of $x$ by the action of $N$.

We denote by $C_G(x) = \{g \in G \mid gx = xg\}$, $C_N(x) = \{n \in N \mid nx = xn\}$, the centralizers of $x$ ($\iff$ the stabilizers of the action by conjugation).

I was asked to prove that if $x \in N$, either $x^N = x^G$ (if $C_G(x) \not\subset N$), or $\lvert x^N \rvert = \frac{1}{2} \lvert x^G \rvert$ (if $C_G(x) \subset N \iff C_G(x) = C_N(x)$). However, I don't understand why we need $x \in N$. Is it wrong in the general case $x \in G$ ?

By Orbit-Stabilizer theorem, we have $[G:C_G(x)] = x^G$ and $[N:C_N(x)] = x^N \ \forall x \in G$. Furthermore, $x^N \subseteq x^G$. Therefore : $$[C_G(x):C_N(x)] = \frac{\lvert C_G(x) \rvert}{\lvert C_N(x) \rvert} = \frac{\frac{\lvert G \rvert}{\lvert x^G \rvert}}{\frac{\lvert N \rvert}{\lvert x^N \rvert}} = 2 \frac{\lvert x^N \rvert}{\lvert x^G \rvert} \leq 2$$

so we deduce the result from that. But here, I didn't use $x \in N$, so I don't know why it was in hypotheses. Have I done something wrong ?

  • 2
    No you don't need $x \in N$ for this result to hold, although if $x \not\in N$ then you always have $x^N=x^G$ - do you see why? – 2017-01-26
  • 0
    @DerekHolt I'm clueless here :/ – 2017-01-26
  • 1
    Well if $x \not\in N$ then clearly $C_G(x)$ is not contained in $N$. – 2017-01-27
  • 0
    I thought about this but it didn't seem that clear to me. But well, taking the contrapositive is easier. If $C_G(x) \subseteq N$, then $x \in N$ is kind of obvious. Thank you. – 2017-01-27

0 Answers 0