- #1

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

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- Thread starter Mr Davis 97
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- #1

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

- #2

fresh_42

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

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

- #3

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Then why does the exercise bother with specifying that ##N## should be finite?Sure. It is even true for any subset ##N## ... as long as the quantor at ##g## is an ##\forall##.

- #4

fresh_42

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

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##.

- #5

Stephen Tashi

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- #6

fresh_42

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So the assumption

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?

- #7

martinbn

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

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