Math Help Forum - View Single Post - Group Theory

tonio · #4 November 5th, 2009, 04:34 AM

Quote:

Originally Posted by Swlabr

Can we not just apply the Burnside Basis Theorem and prove that all groups of order $p^2$ are Abelian? This would clearly prove the result.

Let $G$ be a finite $p$ -group, and denote by $G'$ the derived subgroup, $G'=<[g,h]:g,h \in G>=<g^{-1}h^{-1}gh:g,h \in G>$ .

Denote by $\Phi(G)$ the Frattini subgroup; the intersection of all the maximal subgroups of $G$ . Then $G/\Phi(G)$ is an elementary Abelian group which we can view as a vector space of rank $d(G)$ . The Burnside Basis Theorem states:

i) $\Phi(G)=G'G^p$ .

ii) For $S \subseteq G$ such that $<S>=G$ then there exists $T \subseteq S$ such that $|T|=d(G)$ and $<T>=G$ .

iii) Another statement that isn't really relevant here but basically says that the generators of the group are precisely the coset representatives for the generators of the quotient group.

Thus, if $|\Phi(G)|=p$ we have that the group is cyclic (by part (ii) ), and if $|\Phi(G)|=1$ then we have that $G'=1$ and the group is Abelian (by part (i) ). Clearly there exists a subgroup of order $p$ in the group so the group cannot have $\Phi(G)=G$ . Thus the group is Abelian.

I know it is a bit like overkill, but it is my favorite theorem...

The OP stated that the problem is given before the class equation, Sylow's theorems and stuff, so imagine Frattini's group!
Your point (ii) there though looks a little odd: it simply says that from a given set of generators of G, a finite group, you can always take a subset which equals the rank of the group=the minimal number of elements required to generate the group, which seems to be false (but I haven't given this too much thought), unless you meant that $T=\subset S\setminus \Phi(G)$ , and anyway I don't see how from this it follows that if $|\Phi(G)|=p$ then G is cyclic...

Tonio