Why must the group N be finite in this result?

  • #1
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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?
 

Answers and Replies

  • #2
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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##.
 
  • #3
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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?
 
  • #4
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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##.
 
  • #6
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  • #7
martinbn
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##gNg^{-1}## and ##N## have the same index.
 

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