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Topic: Abelian or nontrivial intersection? (Read 953 times) 

Michael Dagg
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Abelian or nontrivial intersection?
« on: Jan 6^{th}, 2007, 6:33pm » 
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Let N be a normal subgroup of G such that G = H x K. Show that N must be abelian or intersects one of the factors H or K nontrivially.


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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #1 on: Jan 19^{th}, 2007, 4:49pm » 
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M_D squeezes unexpected blood from direct products of groups. Here is a hint. The maps f(n), from N into H, and g(n), from N into K, defined by n=f(n)g(n), for each n in N, are homomorphisms of N. After exploring relevant consequences of the properties of f and g, the problem reduces to one short calculation, where the normality of N finally comes into play.


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Eigenray
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Re: Abelian or nontrivial intersection?
« Reply #2 on: Jan 20^{th}, 2007, 10:55am » 
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If N is nonabelian, then there are hk, h'k' in N such that [hk, h'k'] = [h,h'][k,k'] is nontrivial. WLOG then, [h,h'] is nontrivial. Since N is normal, it contains [hk, h'] = [h,h'][k,1] = [h,h'], as desired. A similar argument will show that if S is nonabelian simple, then any normal subgroup of S^{n} is a product of some subset of the factors, and then that Aut(S^{n}) = Aut(S)^{n} x S_{n}.


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Michael Dagg
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Re: Abelian or nontrivial intersection?
« Reply #3 on: Mar 12^{th}, 2007, 6:24pm » 
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Nice! Ecoist wrote a nice solution to this problem that a refects his remark above. Maybe he will post it. (I am seeing that the email notification of postings does not always work.)


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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #4 on: Mar 12^{th}, 2007, 7:43pm » 
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I did not post my solution because I think Eigenray's solution is more efficient. I prefer proofs that use as few weapons as possible, providing a more accurate assessment of the depth of the problem.


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Michael Dagg
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Re: Abelian or nontrivial intersection?
« Reply #5 on: Mar 12^{th}, 2007, 8:36pm » 
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What you wrote is a textbook solution  that is, one that clearly complements material from which a problem like this can be drawn (i.e. your homomorphic characterization of f and g include several things that are closely accessible and then lead to another problem like this one). Weaponry? I thought if you had it then everyone (we) should see it.

« Last Edit: Mar 12^{th}, 2007, 8:38pm by Michael Dagg » 
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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #6 on: Mar 12^{th}, 2007, 9:23pm » 
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Ok. Assuming my proof has some value, I will post my solution later (got first round of golf for the year in the morning!).


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Michael Dagg
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Re: Abelian or nontrivial intersection?
« Reply #7 on: Mar 12^{th}, 2007, 9:32pm » 
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So, you have a putt...


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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #8 on: Mar 13^{th}, 2007, 9:14pm » 
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In light of Eigenray's proof, the proof below is somewhat embarassing. But I respect Michael_Dagg. Define maps f and g from N into, resp., H and K by, for each n in N, f(n) is the unique element in H and g(n) is the unique element in K such that n=f(n)g(n). Then f is a homomorphism from N into H and g is a homomorphism from N into K. If f or g is not an isomorphism, then N meets H or K nontrivially. If N is not abelian, then f(N) is not abelian. Hence, for some h and h' in f(N), we have h^{1}h'hh'^{1} is not 1. Let n'=h'g(n'). We compute h^{1}n'hn'^{1}. h^{1}n'hn'^{1} = h^{1}h'g(n')hh'^{1}g(n')^{1} = h^{1}h'hh'^{1}. This nonidentity element lies in H, and also in N because N is normal in G. Hence N intersects H nontrivially.


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