If p and q are primes then every group of order p^m q^n is solvable.

Feb 2009
189
0
Hi: Burnside proved (using representation theory) that the number of elements in a conjugacy class of a finite simple group can never be a prime power larger than 1. Use this fact to prove Burnside's theorem: If p and q are primes then every group of order p^m q^n is solvable.

What I did: let |G|= p^m q^n and suppose G simple. Let T(g) be the conjugacy class of some element g in G. Of course |Z(G)|= 1. If |T(g)| != 1 then pq | |T(g)|, for T(g) cant be a prime power and T(g) | |G|. Therefor the class equation is |G| = 1 + c, where p|c and hence p | 1, which is absurd. Therefor G has a nontrivial proper normal subgroup H. (1)

If |G|=1 then G is vacuously solvable [?]. Assume the statement in Burnside'a theorem true for every group of order less than |G|. By (1) G has a nontrivial proper normal subgroup H. Now |H| < |G| and |G/H| < |G|. By the induction hypothesis H and G/H are solvable. According to a previous theorem then G is solvable.

The question: is this a valid proof by mathematical induction? For certainly |G| does not take all integer values less than |G|. In other words, shouldn't I do induction on m, or induction on both m and n?
 
Dec 2012
1,145
502
Athens, OH, USA
No, your proof is not valid.
What I did: let |G|= p^m q^n and suppose G simple. Let T(g) be the conjugacy class of some element g in G. Of course |Z(G)|= 1. If |T(g)| != 1 then pq | |T(g)|, for T(g) cant be a prime power and T(g) | |G|. Therefor the class equation is |G| = 1 + c, where p|c and hence p | 1, which is absurd. Therefor G has a nontrivial proper normal subgroup H. (1)
You can not conclude that pq divides the order of the conjugacy class T(g) of any non-central element g. According to Burnside's theorem, the order of T(g) is p or q or $p^aq^b$ with both a and b greater than or equal to 1.
In fact, if G is a simple group of order $p^mq^n$, G has non_central elements $g_1$ and $g_2$ with $|T(g_1)|=p$ and $|T(g_2)|=q$. You prove this. Now argue further to conclude G is not simple.

Your second paragraph is fine.
 
Dec 2012
1,145
502
Athens, OH, USA
Sorry, my previous post is wrong. I misinterpreted your statement of Burnside's theorem. I read (or reread; I've forgotten if I ever knew the theorem) and saw that the statement is that if a finite group G of order greater than 1 has any conjugacy class of prime power order, then G is not simple. So your proof is fine.

I was thinking that in a group G of order $p^mq^n$ an element in the center of a Sylow p subgroup has centralizer of index a power of $q$. By Burnside's theorem, this says G is not simple.
 
Feb 2009
189
0
Well in fact there's something not quite clear in my proof. When making the induction, if |H| happens to be p^n then I can't use the induction hypothesis, because |H| will then be q^n. However, if |H| = p^n then H is a p-group and hence solvable as well as G/H.