0

I want to show that for some proper subgroup $H\subsetneq G$ ($G$ finite) we always have $$G\neq\bigcup_{g\in G}gHg^{-1}$$ My attempt so far is letting $G$ act on the set $M$ of its subgroups of order $|H|$ by conjugation: $$\begin{matrix}G.M\end{matrix}\to M\\g.H\mapsto H^g$$

The stabilizer $G_H$ obviously then contains $H$, so $G.H$ is not empty and therefor we have that $|G.H|$ is a non trivial divisor of $|G|$. How do I continue from here?

Stefan4024
  • 35,843
Buh
  • 1,395
  • 7
  • 10

1 Answers1

1

HINT: Prove that there are exacty $[G:N[H]] \le [G:H]$ distinct subsets on the right. If the equality holds the intersection of any two must be an empty set. However the identity is in all of them.

$$gHg^{-1} = kHk^{-1} \iff (k^{-1}g)H(g^{-1}k) = H \iff k^{-1}g \in N[H] \iff gN[H] = kN[H]$$ This helps you conclude that the number of distinct subsets $gHg^{-1}$ as $g$ runs through $G$ is same as the number of distinct cosets of $N[H]$. Finally use the fact $H \le N[H]$

Stefan4024
  • 35,843
  • You probably mean that by applying Lagrange's theorem twice to $H\subset N(H)\subset G$ we have that $[G:H]=[G:N(H)]\cdot[N(H):H]$ – Buh May 10 '18 at 08:27
  • @Buh Yeah, that will work – Stefan4024 May 10 '18 at 08:28
  • But then I don't really understand why I would need the identity to get the contradiction. Don't we just have that $|\bigcup_{g\in G}gHg^{-1}|=[G:N(H)]$ by your argument in the hidden box where $[G:N(H)]\neq |G|$? – Buh May 10 '18 at 08:37
  • You have $[G:N(H)]$ sets on the right with size |H|. On the other side G can be written as $[G:H]$ sets of size |H| by cosets. If the equality holds then the distinct sets on the right has to have empty intersection. – Stefan4024 May 10 '18 at 08:47
  • Alright, thanks! – Buh May 10 '18 at 09:05