Frattini subgroup contained in center implies derived subgroup is elementary abelian

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Statement

Suppose P is a group of prime power order: in other words, P has order p^n for some prime p and integer n. Further, suppose that the Frattini subgroup of P is contained in the center of P; in symbols:

\Phi(P) \le Z(P)

Then the commutator subgroup (or derived subgroup) of P is elementary Abelian.

Facts used

  1. For a group of prime power order, any subgroup containing the Frattini subgroup, has a quotient which is an elementary Abelian group
  2. For a group of prime power order, the derived subgroup is contained in the Frattini subgroup. For full proof, refer: Nilpotent implies derived in Frattini
  3. In a group of nilpotence class two, the map y \mapsto [x,y], for fixed x, is an endomorphism. For full proof, refer: Class two implies commutator map is endomorphism

Proof

P' \le \Phi(P), and since \Phi(P) \le Z(P), and Z(P) is Abelian, P' is Abelian. Hence, to show that it is elementary Abelian, it suffices to show that it is generated by elements of order p.

We know that P' is generated by commutators, i.e. elements of the form [x,y] where x,y \in P, so it suffices to show that any commutator has order P.

Let's do this. Since P' \le Z(P), the group P has nilpotence class two. By fact (2), the map y \mapsto [x,y] is, for fixed x, an endomorphism of P. Thus, we have:

[x,y^p] = [x,y]^p

Z(P) contains \Phi(P), so by fact (1), the quotient P/Z(P) is elementary Abelian. Equivalently, for any y \in P, the element y^p \in Z(P). Thus, the left side in the above equation is the identity element, showing that forany x,y \in G, [x,y]^p = e, completing the proof.