Solvable groups

  • Thread starter bham10246
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  • #1
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I've been working on this problem and I need just a small hint.

Let [itex]A[/itex] and [itex]B[/itex] be solvable subgroups of a group [itex]G[/itex] and suppose that [itex]A\triangleleft G [/itex]. Prove that [itex]AB[/itex] is solvable.


My idea:

So we have a chain of normal subgroups of A so that their quotient is abelian. We also have a chain of normal subgroups of B so that their quotient is abelian. Since A is normal in G, should I multiply the normal subgroups [itex]A_i[/itex] in A by B to obtain [itex]B=1*B=A_0 B \triangleleft A_1 B \triangleleft ... \triangleleft A_k B = AB[/itex], but how do we know that [itex](A_{i+1}B)/(A_i B)[/itex] is abelian?

If I understand this one thing, then I think I can finish the rest of the proof. Thank you!

This is a right approach, right?
 

Answers and Replies

  • #2
mathwonk
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somehow i am tempted to use the fundamental isomorphism theorem about

the structure of (AB)/A.
 
  • #3
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Thanks, I'll try that. I thought that the above construction [itex]B=1*B=A_0 B \triangleleft A_1 B \triangleleft ... \triangleleft A_k B = AB[/itex] is correct but I'm not even sure that [itex]A_i B[/itex] is normal in [itex]A_{i+1}B[/itex]?!
 

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