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The Frattini subgroups of the dihedral and quaternion groups of order eight

Example

Let D:=Dih(C4)=D4 in the convention of Dih(Cn)=CnC2 with inversion action has order 2n and the dihedral relations, so D is the dihedral group of order eight, and let Q8 be the quaternion group. Then

Φ(D)=r2,Φ(Q8)={1,1}.

For the dihedral group of order eight and Q8, the Frattini subgroup has order two and the Frattini quotient is (Z/2)2. Both groups have generator rank two.

Facts & Assumptions

[L1]

For every finite 2-group P, Φ(P)=P2 (Φ(P)=P2 for a finite 2-group).

[L2]

In D=Dih(C4), r4=s2=1, srs1=r1, and every element is ri or ris; in Q8, the elements ±i,±j,±k have order four and 1 is the unique element of order two ( Dih(Cn)=CnC2 with inversion action has order 2n and the dihedral relations, Q8 is a subgroup of H× with eight elements, and 1 is its only element of order 2).

[F1]

The commutator subgroup is generated by [g,h]=ghg1h1 (Commutators [g,h]=ghg1h1 and the commutator subgroup [G,G]).

[F2]

The generator rank is the common size of a basis of the Frattini quotient (The generator rank d(P) of a finite p-group).

Verification

technique · direct
1.1

In D, (ri)2=r2i and (ris)2=1, so D2=r2. In Q8, the squares are 1 and 1, with every noncentral element squaring to 1, so Q82={1,1}. The commutators of [F1] give the same two subgroups: [r,s]=r(sr1s1)=rr=r2 and every commutator of D is a power of r2, so D=r2; and [i,j]=iji1j1=k(i)(j)=kij=k2=1, so Q8={1,1}.

givenL2F1algebra
2.1

Apply [L1] to step 1.1. Each quotient has order four and exponent two, with the classes of r,s and of i,j respectively as two-vector bases. Thus both quotients are (Z/2)2, and [F2] gives generator rank two.

step 1.1L1L2F2algebra

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