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        riddles >> putnam exam (pure math) >> Unique subgroup of a finite group
        
(Message started by: Michael_Dagg on Jan 4^th, 2007, 9:21pm)

Title: Unique subgroup of a finite group
Post by Michael_Dagg on Jan 4^th, 2007, 9:21pm

Suppose H is a normal subgroup of a finite group G such
that (|H|,|G:H|) = 1.

Is H the unique subgroup of G having order |H| ?

Title: Re: Unique subgroup of a finite group
Post by ecoist on May 14^th, 2007, 8:44pm

Don't know why this problem has gone so long without a posted solution. Just thought of an approach that differs from my first (number-theoretic) idea for a solution. What about using the following result?

Let H be a subgroup of the finite group G which contains the normalizer N(P) of a Sylow p-subgroup P of G. Then H is its own normalizer in G.

Pardon me for not posting a solution, but I don't want to spoil things for those for whom group theory is a new and fascinating subject.

Title: Re: Unique subgroup of a finite group
Post by Eigenray on May 14^th, 2007, 10:35pm

My first thought was that this is "obvious" if G is solvable (Hall), but I didn't think about the general case very much. But then I saw the word "Sylow" and it just clicked:

Let K http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif G with |K|=|H|. If p | |H|, then Sylow p-subgroups P,P' of H,K are also Sylow p-subgroups of G, so they are conjugate in G. But since H is normal, we must have P' http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif K http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gif H. Since this holds for all such p, we have |H| | |K http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gif H|, hence K=H.

What did group theorists do before Sylow?

Title: Re: Unique subgroup of a finite group
Post by ecoist on May 15^th, 2007, 3:16pm

As usual, Eigenray's solution is the best, but consider the following equally short solution as well.

Let H have order n and let k=[G:H]. Let x be any element of G of order dividing n. In the factor group G/H, x^k=1 (mod H) and, since x has order dividing n in G, xⁿ=1 (mod H). Since there exist integers u and v such that nu+kv=1, we have

x¹=(xⁿ)^u.(x^k)^v)=1 (mod H).

Hence x lies in H; whence H is the unique subgroup of order n in G.

Title: Re: Unique subgroup of a finite group
Post by Eigenray on May 16^th, 2007, 1:29am

Actually I like yours better. It shows that

H = {x | xⁿ = 1}.

As a followup: Show that G is a Frobenius group, with Frobenius kernel H, iff xⁿ=1 or x^k=1 for all x in G.

Title: Re: Unique subgroup of a finite group
Post by Michael_Dagg on May 18^th, 2007, 12:22pm

Neat solutions!