Group generation and normal subgroups

  • Thread starter Bleys
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There is a step in some book which is vaguely explained, I just want to check whether my working is correct.
Let G be a group, and let x be in G-{1}. Let Y be the set [itex] \left\{g^{-1}xg : g\in G\right\} [/itex].
I want to show <Y> is normal in G. Now it's clear why Y is invariant under conjugation (Y being itself a conjugacy class, namely of x). The book says that from this it follows <Y> is normal, but I didn't really think it that was clear. Why does it follow that also <Y> is invariant under conjugation?

This is what I've done, but I don't know if it's correct (or if it's even the argument the book is implying):
any y in <Y> is of the form [itex]g^{-1}x^{k}g[/itex] where k is a positive integer, or [itex]gx^{k}g^{-1}[/itex] where k is a negative integer. I'll do k=2;the rest follows by induction.
First case: [itex] g^{-1}g^{-1}x^{2}gg = g^{-1}(g^{-1}xg)(g^{-1}xg)g = g^{-1}(g^{-1}xg)gg^{-1}(g^{-1}xg)gg [/itex]Since Y is invariant under conjugation, this is a product of two elements in Y hence in <Y>.
Second case: [itex]g^{-1}gx^{-2}g^{-1}g = x^{-2}[/itex], which is clearly in <Y> since x is in Y.
Then for any y in <Y>, [itex]g^{-1}yg[/itex] is in <Y>.
Is this correct? Is there a simpler way to do it?

EDIT: I don't know why the code is not showing for the First case...
 
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Answers and Replies

  • #2
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It's because [tex]g^{-1}(ab)g=(g^{-1}ag)(g^{-1}bg)[/tex], i.e. conjugation distributes over products, so you only need to verify invariance under conjugation for a set of generators.
 
  • #3
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oh ok; thanks Tinyboss!
 

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