Frattini-in-center p-group implies derived subgroup is elementary abelian

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Suppose P is a group of prime power order with the property that P/Z(P) is elementary Abelian, or equivalently, that \Phi(P) is contained in Z(P) (i.e., P is a Frattini-in-center group). Then, the commutator subgroup P' = [P,P] is an elementary Abelian group: it is an Abelian group of exponent p.

Related facts


Facts used

  1. Class two implies commutator map is endomorphism


Given: A finite p-group P such that P/Z(P) is elementary Abelian.

To prove: P' is elementary Abelian.

Proof: First, since P/Z(P) is elementary Abelian, P' \le Z(P), so P' is in the center. In particular, P' is an Abelian group. Thus, it suffices to show that P' has exponent p.

Note that since P' \le Z(P), P has nilpotence class two. Thus, by fact (1), we have that for any x \in P, the map y \mapsto [x,y] is an endomorphism. In particular, we have that for any x,y \in P:

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

Now, since P/Z(P) is elementary Abelian, y^p \in Z(P), so the left side is the identity element. Thus, [x,y]^p is the identity element for any x,y \in P, and so P' is generated by elements of order p. Since P' is Abelian, this tells us that P' has exponent p.