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Topic: Center of a finite group (Read 2106 times) |
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Michael Dagg
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Center of a finite group
« on: Mar 15th, 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 .
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| « Last Edit: Mar 15th, 2007, 10:05pm by Michael Dagg » |
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Michael Dagg
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Aryabhatta
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Re: Center of a finite group
« Reply #1 on: Mar 16th, 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 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. 
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| « Last Edit: Mar 16th, 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 21st, 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 p-1. |
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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).
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Eigenray
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Re: Center of a finite group
« Reply #3 on: Mar 21st, 2007, 8:55pm » |
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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 . |
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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? |
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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. |
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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.
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Eigenray
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Re: Center of a finite group
« Reply #4 on: Mar 21st, 2007, 9:11pm » |
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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.
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| « Last Edit: Mar 21st, 2007, 9:13pm by Eigenray » |
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Eigenray
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Re: Center of a finite group
« Reply #5 on: Mar 21st, 2007, 9:23pm » |
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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.
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| « Last Edit: Mar 21st, 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 18th, 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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Michael Dagg
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Re: Center of a finite group
« Reply #7 on: May 21st, 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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Michael Dagg
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