Author |
Topic: Center of a finite group (Read 2100 times) |
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Center of a finite group
« on: Mar 15th, 2007, 10:05pm » |
Quote Modify
|
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 15th, 2007, 10:05pm by Michael Dagg » |
IP Logged |
Regards, Michael Dagg
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Center of a finite group
« Reply #1 on: Mar 16th, 2007, 12:20am » |
Quote Modify
|
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 Cz. All we need to show is |Cz| = 1. Suppose Cz =/= {z} (otherwise there is nothing to prove) Let y \in Cz 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 |Cz| can never exceed p-1. But, |Cz| divides |G|, hence p is not the smallest prime which divides |G|, contradiction. Thus we must have that |Cz| = 1. I am pretty sure I am making a basic mistake somewhere.
|
« Last Edit: Mar 16th, 2007, 1:34am by Aryabhatta » |
IP Logged |
|
|
|
Pietro K.C.
Full Member
Gender:
Posts: 213
|
|
Re: Center of a finite group
« Reply #2 on: Mar 21st, 2007, 6:38pm » |
Quote Modify
|
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 p-1. |
| If z has order p, then there are p distinct powers of z; whence the conclusion that #Cz cannot exceed p-1? I took a somewhat different approach. Since z has order p, the cyclic subgroup it generates is (z) = {1, z, ... , zp-1} 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 Zp there is n in Zp such that gzmg-1 = zn. For each m in Zp, denote this n by f(m); it is unique because a,b < p --> [ za = zb --> a = b ]. By a similar argument, one establishes that f is injective; therefore f : Zp --> Zp is a bijection on Zp. (Note that it depends on the choice of g.) Even more is true: gzm+ng-1 = gzmg-1 gzng-1 = zf(m)zf(n) = zf(m)+f(n) so f is in fact an automorphism of Zp 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 fm = 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 Zp is the identity precisely when it maps 1 to 1: fm(1) = fm-1(f(1)) = f(1) fm-1(1) = ... = (f(1))m = 1. Since f(1) is in the multiplicative group Zp* of order p-1, then the least such m will be a divisor of p-1, and therefore f p-1 = id. Now, this gives us an idea. How can we get the (p-1)-iteration of f on an exponent? Well, we can conjugate by g p-1 times: gp-1zg-(p-1) = zid(1) = z whence gp-1z = zgp-1 That is, z commutes with every (p-1)-th power in the group; but p being the smallest prime factor of #G, p-1 is prime with #G, and the function x |-> xp-1 is injective, hence bijective, since G is finite. Therefore, z is in Z(G).
|
|
IP Logged |
"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Center of a finite group
« Reply #3 on: Mar 21st, 2007, 8:55pm » |
Quote Modify
|
on Mar 15th, 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 non-abelian, it would be Hamiltonian, but Q8 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 21st, 2007, 6:38pm, Pietro K.C. wrote:If z has order p, then there are p distinct powers of z; whence the conclusion that #Cz cannot exceed p-1? |
| Any element conjugate to z must be a non-trivial power of z: <z> has order p, but only p-1 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 p-1, 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.
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Center of a finite group
« Reply #4 on: Mar 21st, 2007, 9:11pm » |
Quote Modify
|
on Mar 21st, 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 21st, 2007, 9:13pm by Eigenray » |
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Center of a finite group
« Reply #5 on: Mar 21st, 2007, 9:23pm » |
Quote Modify
|
Another proof: we can easily prove by induction that any finite p-group satisfying (*) is cyclic. So if G satisfies (*), all its Sylow subgroups are cyclic, so by Holder-Burnside-Zassenhaus, G is a semidirect product of relatively prime order cyclic groups, but this product must be direct, so G is cyclic.
|
« Last Edit: Mar 21st, 2007, 9:24pm by Eigenray » |
IP Logged |
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Center of a finite group
« Reply #6 on: May 18th, 2007, 12:23pm » |
Quote Modify
|
> Actually, is this where you were going with this problem? Sure was.
|
|
IP Logged |
Regards, Michael Dagg
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Center of a finite group
« Reply #7 on: May 21st, 2007, 9:34pm » |
Quote Modify
|
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).
|
|
IP Logged |
Regards, Michael Dagg
|
|
|
|