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Topic: Abelian or nontrivial intersection? (Read 955 times) |
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Michael Dagg
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Abelian or nontrivial intersection?
« on: Jan 6th, 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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Regards,
Michael Dagg
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ecoist
Senior Riddler
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Re: Abelian or nontrivial intersection?
« Reply #1 on: Jan 19th, 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
wu::riddles Moderator
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Re: Abelian or nontrivial intersection?
« Reply #2 on: Jan 20th, 2007, 10:55am » |
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If N is non-abelian, then there are hk, h'k' in N such that [hk, h'k'] = [h,h'][k,k'] is non-trivial. WLOG then, [h,h'] is non-trivial. 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 non-abelian simple, then any normal subgroup of Sn is a product of some subset of the factors, and then that Aut(Sn) = Aut(S)n x| Sn.
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Michael Dagg
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Re: Abelian or nontrivial intersection?
« Reply #3 on: Mar 12th, 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 e-mail notification of postings does not
always work.)
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Michael Dagg
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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #4 on: Mar 12th, 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 12th, 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.
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| « Last Edit: Mar 12th, 2007, 8:38pm by Michael Dagg » |
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Michael Dagg
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ecoist
Senior Riddler
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Re: Abelian or nontrivial intersection?
« Reply #6 on: Mar 12th, 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
Senior Riddler
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Re: Abelian or nontrivial intersection?
« Reply #7 on: Mar 12th, 2007, 9:32pm » |
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So, you have a putt...
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Regards,
Michael Dagg
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ecoist
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Re: Abelian or nontrivial intersection?
« Reply #8 on: Mar 13th, 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-1h'hh'-1 is not 1. Let n'=h'g(n'). We compute h-1n'hn'-1.
h-1n'hn'-1 = h-1h'g(n')hh'-1g(n')-1 = h-1h'hh'-1.
This non-identity element lies in H, and also in N because N is normal in G. Hence N intersects H nontrivially.
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