Order has only two prime factors implies solvable

From Groupprops

Jump to: navigation, search

Contents

Statement

The Burnside's p^aq^b theorem, states that a group whose order has at most two prime factors, (viz it is of the form paqb where p,q are primes and a,b are nonnegative integers), then the group must be solvable.

Proof

The overall plan is:

  • Obtain a proper nontrivial normal subgroup
  • Then, use induction on the normal subgroup and quotient group

There exists a conjugacy class of prime power order

Let the order of the group G be paqb where p,q are the prime divisors of the order. Let P be a p-Sylow subgroup, and let Z(P) be the center of P. Z(P) is a nontrivial group. Now for any nonidentity element g \in Z(P), CG(g) contains P, so its index in G is a power of q.

There are now two possibilities:

  • CG(g) = G, so g \in Z(G). Thus, the center is a nontrivial subgroup. If the center is the whole of G, G is an Abelian group, and there is nothing to prove. Otherwise, we have obtained a proper nontrivial normal subgroup.
  • Otherwise, CG(g) is a proper subgroup of G, and thus the conjugacy class of g has size a power of q that is strictly greater than 1.

Existence of such a conjugacy class implies solvability

Further information: (Conjugacy class of prime power order) implies not simple

In the page linked to above, it is shown that if such a conjugacy class exists inside the group, then the group is not simple.

Once we have shown that it is not simple

G now has a proper nontrivial normal subgroup N. Both N and G / N have strictly smaller order than G and both of them have one or two prime divisors. Thus, by induction, both N and G / N are solvable groups, and hence, so is G.

Retrieved from "http://groupprops.subwiki.org/wiki/Order_has_only_two_prime_factors_implies_solvable"
Personal tools