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Topic: Unique subgroup of a finite group (Read 1041 times) |
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Michael Dagg
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Unique subgroup of a finite group
« on: Jan 4th, 2007, 9:21pm » |
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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| ?
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| « Last Edit: Jan 4th, 2007, 10:41pm by Michael Dagg » |
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Michael Dagg
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ecoist
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Re: Unique subgroup of a finite group
« Reply #1 on: May 14th, 2007, 8:44pm » |
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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.
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Eigenray
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Re: Unique subgroup of a finite group
« Reply #2 on: May 14th, 2007, 10:35pm » |
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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  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' K  H. Since this holds for all such p, we have |H| | |K H|, hence K=H.
What did group theorists do before Sylow?
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ecoist
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Re: Unique subgroup of a finite group
« Reply #3 on: May 15th, 2007, 3:16pm » |
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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, xk=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.(xk)v)=1 (mod H).
Hence x lies in H; whence H is the unique subgroup of order n in G.
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Eigenray
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Re: Unique subgroup of a finite group
« Reply #4 on: May 16th, 2007, 1:29am » |
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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 xk=1 for all x in G.
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Michael Dagg
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Re: Unique subgroup of a finite group
« Reply #5 on: May 18th, 2007, 12:22pm » |
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Neat solutions!
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Michael Dagg
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