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Topic: Center of a finite group (Read 2093 times) 

Michael Dagg
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Center of a finite group
« on: Mar 15^{th}, 2007, 10:05pm » 
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Let G be a finite group such that, for each positive integer n dividing the order of G, there is at most one subgroup of G of order n . Suppose p is the smallest prime dividing the order of G and let z in G be an element of order p. Show that z is in the center of G .

« Last Edit: Mar 15^{th}, 2007, 10:05pm by Michael Dagg » 
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Aryabhatta
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Re: Center of a finite group
« Reply #1 on: Mar 16^{th}, 2007, 12:20am » 
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My group theory is rusty, but I will give it a shot. Let z be an element of order p (where p is the smallest prime) Consider the conjugacy class C_{z}. All we need to show is C_{z} = 1. Suppose C_{z} =/= {z} (otherwise there is nothing to prove) Let y \in C_{z} Cleary y^p = e (as z^p = e) Consider the cyclic groups <z> and <y>. Since p is the smallest prime which divides G, we must have that <z> = <y> and hence by the problem hypothesis, <z> = <y> Thus y is a power of x. Thus we see that C_{z} can never exceed p1. But, C_{z} divides G, hence p is not the smallest prime which divides G, contradiction. Thus we must have that C_{z} = 1. I am pretty sure I am making a basic mistake somewhere.

« Last Edit: Mar 16^{th}, 2007, 1:34am by Aryabhatta » 
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Pietro K.C.
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Re: Center of a finite group
« Reply #2 on: Mar 21^{st}, 2007, 6:38pm » 
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A nice problem! Aryabhatta's solution starts out very well, but I'm not sure about the following passage: Quote: Thus y is a power of z. Thus we see that Cz can never exceed p1. 
 If z has order p, then there are p distinct powers of z; whence the conclusion that #C_{z} cannot exceed p1? I took a somewhat different approach. Since z has order p, the cyclic subgroup it generates is (z) = {1, z, ... , z^{p1}} By hypothesis, this is the only subgroup of order p, so (z) is a normal subgroup. (Briefly, for any g in G, one has g(z)g^{1} a subgroup of same cardinality, hence the same subgroup.) This means that, for any g in G and u in (z), there is another v in (z) such that gug^{1} = v. Looking at (z), we can further specify that, for any m in Z_{p} there is n in Z_{p} such that gz^{m}g^{1} = z^{n}. For each m in Z_{p}, denote this n by f(m); it is unique because a,b < p > [ z^{a} = z^{b} > a = b ]. By a similar argument, one establishes that f is injective; therefore f : Z_{p} > Z_{p} is a bijection on Z_{p}. (Note that it depends on the choice of g.) Even more is true: gz^{m+n}g^{1} = gz^{m}g^{1} gz^{n}g^{1} = z^{f(m)}z^{f(n)} = z^{f(m)+f(n)} so f is in fact an automorphism of Z_{p} as a +group. Since all of its elements are sums of 1's, f(mn) = m f(n). Now we inquire into the cycle structure of f: is there a natural number m such that f^{m} = id? (Superscripts on functions denote iteration; on field elements, exponentiation.) Well, since its domain is finite and it is a bijection, there must be; but we may use its particular structure to determine this m more precisely. An automorphism of Z_{p} is the identity precisely when it maps 1 to 1: f^{m}(1) = f^{m1}(f(1)) = f(1) f^{m1}(1) = ... = (f(1))^{m} = 1. Since f(1) is in the multiplicative group Z_{p}* of order p1, then the least such m will be a divisor of p1, and therefore f ^{p1} = id. Now, this gives us an idea. How can we get the (p1)iteration of f on an exponent? Well, we can conjugate by g p1 times: g^{p1}zg^{(p1)} = z^{id(1)} = z whence g^{p1}z = zg^{p1} That is, z commutes with every (p1)th power in the group; but p being the smallest prime factor of #G, p1 is prime with #G, and the function x > x^{p1} is injective, hence bijective, since G is finite. Therefore, z is in Z(G).


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Eigenray
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Re: Center of a finite group
« Reply #3 on: Mar 21^{st}, 2007, 8:55pm » 
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on Mar 15^{th}, 2007, 10:05pm, Michael_Dagg wrote:Let G be a finite group such that, for each positive integer n dividing the order of G, there is at most one subgroup of G of order n . 
 If G were nonabelian, it would be Hamiltonian, but Q_{8} has 3 subgroups of order 4. So by Baer's theorem, G must be abelian; since it is finite, it must in fact be cyclic. Therefore z is in the center of G. on Mar 21^{st}, 2007, 6:38pm, Pietro K.C. wrote:If z has order p, then there are p distinct powers of z; whence the conclusion that #C_{z} cannot exceed p1? 
 Any element conjugate to z must be a nontrivial power of z: <z> has order p, but only p1 generators. Quote:I took a somewhat different approach. 
 What you are doing here, essentially, is very common in group theory: given a normal subgroup N of a group G, we can consider the homomorphism : G Aut(N), where (g) is the automorphism of N given by conjugation by g, i.e., (g) takes n to gng^{1}. In the case N has order p, Aut(N) has order p1, which is relatively prime to G, and therefore any homomorphism G Aut(N) must be trivial. But this means exactly that N is in the center of G.


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Eigenray
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Re: Center of a finite group
« Reply #4 on: Mar 21^{st}, 2007, 9:11pm » 
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on Mar 21^{st}, 2007, 8:55pm, Eigenray wrote: Actually, is this where you were going with this problem? Suppose G is a finite group satisfying (*) any two subgroups of the same order are equal. We can always pick a subgroup P G of order p, where p is the smallest prime dividing G. Then P Z(G), so G/Z(G) is a strictly smaller group, and we can see that it also satisfies (*). Therefore, by induction, G/Z(G) is cyclic, and by a standard result, G is abelian. And a finite abelian group satisfying (*) must be cyclic.

« Last Edit: Mar 21^{st}, 2007, 9:13pm by Eigenray » 
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Eigenray
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Re: Center of a finite group
« Reply #5 on: Mar 21^{st}, 2007, 9:23pm » 
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Another proof: we can easily prove by induction that any finite pgroup satisfying (*) is cyclic. So if G satisfies (*), all its Sylow subgroups are cyclic, so by HolderBurnsideZassenhaus, G is a semidirect product of relatively prime order cyclic groups, but this product must be direct, so G is cyclic.

« Last Edit: Mar 21^{st}, 2007, 9:24pm by Eigenray » 
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Michael Dagg
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Re: Center of a finite group
« Reply #6 on: May 18^{th}, 2007, 12:23pm » 
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> Actually, is this where you were going with this problem? Sure was.


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Michael Dagg
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Re: Center of a finite group
« Reply #7 on: May 21^{st}, 2007, 9:34pm » 
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My apologies Pietro K.C. you wrote a nice solution here and I am just now reading it in detail. (It reminds me to reference it in another problem here).


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