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The Frattini quotient is the largest elementary abelian quotient of a finite p-group

Statement

For a finite p-group P, the quotient P/Φ(P) is elementary abelian (Elementary abelian p-groups, The quotient group G/N and coset product (gN)(hN)=ghN), and for NP the quotient P/N is elementary abelian if and only if Φ(P)N.

Facts & Assumptions

Given: A finite p-group P, its Frattini subgroup Φ(P), and a normal subgroup NP.

[F1]

For a finite group G, Φ(G) is the intersection of all maximal proper subgroups; for G=1 the intersection is G (The Frattini subgroup Φ(G) as the intersection of the maximal subgroups of a finite group).

[L1]

Every finite p-group is nilpotent; every maximal proper subgroup of a finite nilpotent group is normal and has prime index; Lagrange's theorem makes that index divide P=pn, so the index is p; and every group of prime order is cyclic (Every finite p-group is nilpotent, Maximal subgroups of finite nilpotent groups are normal of prime index, Lagrange's theorem: G=[G:H]H for every subgroup H of a finite group G, A finite group of prime order is cyclic and every nonidentity element generates it).

[L3]

For NG, subgroups of G/N correspond inclusion-preservingly to subgroups of G containing N (Correspondence theorem: subgroups of G/N correspond to subgroups of G containing N, with normality preserved).

[L4]

For every finite group G, Φ(G) is characteristic and hence normal (The Frattini subgroup of a finite group is characteristic).

Proof

technique · direct
1.1

By [L1], each maximal subgroup M of P is normal with P/M cyclic of order p. Hence every commutator and every pth power lies in every M. Their images are therefore trivial modulo the normal subgroup Φ(P) from [L4] and [F1], so P/Φ(P) is abelian of exponent at most p, and thus elementary abelian, including the trivial quotient.

givenF1L1L4algebra
1.2

For the forward direction of the kernel criterion, suppose P/N is elementary abelian and let xN. The nonzero vector xN extends by [L2] to a basis of P/N. The span of the other basis vectors is a maximal proper subgroup not containing xN; by [L3], its preimage is a maximal subgroup of P containing N but not x. Thus xΦ(P), and so Φ(P)N.

givenL2L3algebra
2.1

For the reverse direction, suppose Φ(P)N. Step 1.1 places every commutator and every pth power of P inside Φ(P) and hence inside N, so P/N is abelian and every element of it has pth power the identity. It is therefore elementary abelian, including the trivial quotient N=P. Together with step 1.2 this proves the iff.

step 1.1step 1.2F1algebra

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