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        riddles >> putnam exam (pure math) >> Center of a finite group
        
(Message started by: Michael_Dagg on Mar 15^th, 2007, 10:05pm)

Title: Center of a finite group
Post by Michael_Dagg on Mar 15^th, 2007, 10:05pm

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 .

Title: Re: Center of a finite group
Post by Aryabhatta on Mar 16^th, 2007, 12:20am

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 p-1.
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. :-/

Title: Re: Center of a finite group
Post by Pietro K.C. on Mar 21^st, 2007, 6:38pm

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 #C_z cannot exceed p-1?

I took a somewhat different approach. Since z has order p, the cyclic subgroup it generates is

(z) = {1, z, ... , z^p-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 Z_p there is n in Z_p such that

gz^mg^-1 = zⁿ.

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+ng^-1 = gz^mg^-1 gzⁿg^-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^m-1(f(1)) = f(1) f^m-1(1) = ... = (f(1))^m = 1.

Since f(1) is in the multiplicative group Z_p* 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:

g^p-1zg^-(p-1) = z^id(1) = z

whence

g^p-1z = zg^p-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 |-> x^p-1

is injective, hence bijective, since G is finite. Therefore, z is in Z(G).

Title: Re: Center of a finite group
Post by Eigenray on Mar 21^st, 2007, 8:55pm

on 03/15/07 at 22:05:23, 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 [link=http://planetmath.org/encyclopedia/QuasiHamiltonianGroup.html]Hamiltonian[/link], 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 03/21/07 at 18:38:05, 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 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif: G http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif Aut(N), where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif(g) is the automorphism of N given by conjugation by g, i.e., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif(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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif Aut(N) must be trivial. But this means exactly that N is in the center of G.

Title: Re: Center of a finite group
Post by Eigenray on Mar 21^st, 2007, 9:11pm

on 03/21/07 at 20:55:21, Eigenray wrote:

G must be abelian

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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif G of order p, where p is the smallest prime dividing |G|. Then P http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 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.

Title: Re: Center of a finite group
Post by Eigenray on Mar 21^st, 2007, 9:23pm

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.

Title: Re: Center of a finite group
Post by Michael_Dagg on May 18^th, 2007, 12:23pm

> Actually, is this where you were going with this problem?

Sure was.

Title: Re: Center of a finite group
Post by Michael_Dagg on May 21^st, 2007, 9:34pm

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).