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        riddles >> putnam exam (pure math) >> Abelian or nontrivial intersection?
        
(Message started by: Michael_Dagg on Jan 6^th, 2007, 6:33pm)

Title: Abelian or nontrivial intersection?
Post by Michael_Dagg on Jan 6^th, 2007, 6:33pm

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.

Title: Re: Abelian or nontrivial intersection?
Post by ecoist on Jan 19^th, 2007, 4:49pm

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.

Title: Re: Abelian or nontrivial intersection?
Post by Eigenray on Jan 20^th, 2007, 10:55am

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| S_n.

Title: Re: Abelian or nontrivial intersection?
Post by Michael_Dagg on Mar 12^th, 2007, 6:24pm

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.)

Title: Re: Abelian or nontrivial intersection?
Post by ecoist on Mar 12^th, 2007, 7:43pm

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.

Title: Re: Abelian or nontrivial intersection?
Post by Michael_Dagg on Mar 12^th, 2007, 8:36pm

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.

::)

Title: Re: Abelian or nontrivial intersection?
Post by ecoist on Mar 12^th, 2007, 9:23pm

Ok. Assuming my proof has some value, I will post my solution later (got first round of golf for the year in the morning!).

Title: Re: Abelian or nontrivial intersection?
Post by Michael_Dagg on Mar 12^th, 2007, 9:32pm

So, you have a putt...

Title: Re: Abelian or nontrivial intersection?
Post by ecoist on Mar 13^th, 2007, 9:14pm

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.