Why must the group N be finite in this result?

[Mr Davis 97]

Ffom exercise 27 of Dummite and Foote: Let $N$ be a finite subgroup of $G$. Show that $gNg^{-1}\subseteq N$ if and only if $gNg^{-1} = N$.

Why must the subgroup $N$ be finite? Isn't this result true for subgroups of any size?

 

[fresh_42]

Ffom exercise 27 of Dummite and Foote: Let $N$ be a finite subgroup of $G$. Show that $gNg^{-1}\subseteq N$ if and only if $gNg^{-1} = N$.

Why must the subgroup $N$ be finite? Isn't this result true for subgroups of any size?

Sure. It is even true for any subset $N$ ... as long as the quantor at $g$ is an $\forall$.

 

[Mr Davis 97]

Sure. It is even true for any subset $N$ ... as long as the quantor at $g$ is an $\forall$.

Then why does the exercise bother with specifying that $N$ should be finite?

 

[fresh_42]

Then why does the exercise bother with specifying that $N$ should be finite?

I have no idea. Given $gSg^{-1}\subseteq S$ for all $g\in G$ and a subset $S\subseteq G$, we have especially for $g^{-1}\, : \,g^{-1}S(g^{-1})^{-1} = g^{-1}Sg \subseteq S$ and thus $S \subseteq gSg^{-1} \subseteq S$.

Maybe the rest of the exercise depends on it. Here we only use the all quantor on the elements of $G$ and that they have an inverse. We don't even need $S\subseteq G$ as long as $G$ operates via conjugation on a set $S$.

 

[Stephen Tashi]

I found this: https://math.stackexchange.com/ques...up-strictly-contained-in-the-initial-subgroup

 

[fresh_42]

I found this: https://math.stackexchange.com/ques...up-strictly-contained-in-the-initial-subgroup

So the assumption for all $g$ has been wrong.

For finite $N$ we get a bijection even for a single $g\in G$. I only skimmed the link, but it seems, that the case $|G/N|<\infty$ hasn't been proven. So why is $\exists \, g\in G \, : \,gNg^{-1} \subseteq N \Longrightarrow gNg^{-1}=N$ true for $N \leq G$ of finite index?

 

[martinbn]

$gNg^{-1}$ and $N$ have the same index.