Concepto

In this section, we introduce the Frattini subgroups which are important for finite group theory, especially for finitep-groups.

Definition 4.3.1 Let G be a finite group. If G 6= 1, let Φ(G) be the in- tersection of all maximal subgroups of G. If G= 1, let Φ(G) = 1. We call Φ(G) theFrattini subgroup ofG.

Obviously, Φ(G) char G.

(2) Let G= Zn and n= p1α1· · ·pαss. Then Φ(G) is the unique subgroup

of Gof orderpα1−1

1 · · ·pαss−1.

Solution: (1) As a permutation group, letSnact on the set Ω ={1,2, . . . , n}.

Then, for each j∈Ω the stabilizer (Sn)j is a maximal subgroup and the in-

tersection Tn_j=1(Sn)j = 1.A similar method shows Φ(An) = 1.

(2) Let Zn = hgi and let n = p1α1· · ·pαss. Then all hgpji are maximal and

their intersection is

hgp1_{i ∩ hg}p2_{i ∩ · · · ∩ hg}ps_i=hgp1p2···ps_i,

which is cyclic of order pα1−1

1 · · ·pαss−1. ¤

One can characterize a Frattini subgroup in terms of nongenerators. Definition 4.3.2 An element x ∈ G is called a nongenerator of G if G= hS, xi impliesG=hSi for any subsetS ofG.

For example, 1 is a nongenerator of any group.

Theorem 4.3.1 (1) The Frattini subgroupΦ(G)ofGconsists of all non- generators of G.

(2) For any subset A⊆Φ(G) and any S⊆G, we have G=hS, Ai if and only ifG=hSi.

Proof: Since (2) follows from (1), we prove only (1): Φ(G) ={x |x is a nongenerator ofG}.

“ ⊆” Let x be not a nongenerator of G. Then G=hS, xi but hSi< G for some setS. Take a maximal subgroupM such thathSi ≤M < G. Then G=hS, xi ≤ hM, xi.It implies that x /∈M and thenx /∈Φ(G).

“ ⊇” Suppose x /∈Φ(G). Then, x /∈M for some maximal subgroup M of G. Hence,hM, xi=G. However, hMi=M 6=G. It implies that x is not

a nongenerator ofG. ¤

Note that, in general, it is not truethat H < G⇒Φ(H)<Φ(G).

For example, by Example 4.3.1, Φ(S8) = 1, butS8 has a subgroupZ8 whose Frattini subgroup is isomorphic toZ₄. However, we have the following the- orem.

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Theorem 4.3.2 Let H≤G and NEG. IfN ≤Φ(H), then N ≤Φ(G).

Proof: By contrary, suppose that N Φ(G). Then there is a maximal subgroup M of G such that N M. Hence G = N M by the maximality of M. Thus H = H ∩N M = N(H ∩M). (For the last equality, see Exercise 1.1.5.) SinceN ≤Φ(H), by Theorem 4.3.1(2), we haveH =H∩M, that is H≤M. It follows that N ≤M, a contradiction. ¤

Corollary 4.3.3 If KEG, then Φ(K)≤Φ(G).

Proof: Since KEG and Φ(K) charK, we have Φ(K)EG. The desired

result follows from Theorem 4.3.2. ¤

Recall that ifGis nilpotent then every factor group ofGis also nilpotent. The next theorem shows under what conditions the converse is also true Theorem 4.3.4 (Gasch¨utz) LetDandN be normal subgroups of a group GwithD≤N and letN/D be nilpotent. If D≤Φ(G), then N is nilpotent.

Proof: Let P be a Sylow p-subgroup of N. Then P D/D is a Sylow p- subgroup ofN/D, by Proposition 2.2.3(6). SinceN/D is nilpotent, we have P D/DcharN/D. HenceP D/DEG/D, andP DEG. (See Exercise 1.2.1.) By Frattini argument, we have G =N_G(P)P D. SinceD ≤Φ(G), by The- orem 4.3.1, G= NG(P)P = NG(P), or equivalently P EG, which implies

PEN. By the arbitrariness ofp, every Sylow subgroup ofN is normal, and

henceN is nilpotent. ¤

The following two corollaries are immediate. (Also, one can show directly that every Sylow subgroup of Φ(G) is normal by Frattini argument.) Corollary 4.3.5 (Frattini Theorem) The Frattini subgroup Φ(G) of a finite group G is nilpotent.

Corollary 4.3.6 G is nilpotent if and only ifG/Φ(G) is nilpotent.

Proof: By contrary, suppose that p - |G/Φ(G)|. Then, every Sylow p- subgroup P of Φ(G) is also a Sylow p-subgroup of G. By Corollary 4.3.5, Φ(G) is nilpotent, and henceP char Φ(G) andPEG. Since (|G/P|,|P|) = 1, by Schur-Zassenhaus Theorem, P has a complement H in G. Hence G = HP. SinceP ≤Φ(G), we haveG=H, a contradiction. ¤

Now, a nilpotent group can be characterized by its Frattini subgroup. Theorem 4.3.8 (Wielandt) A finite group G is nilpotent if and only if G0 ≤Φ(G).

Proof: If G0 _{≤ Φ(G), then G/Φ(G) is abelian and hence nilpotent. It} follows from Corollary 4.3.6 thatGis nilpotent. Conversely, ifGis nilpotent, then every maximal subgroupM ofGhas index|G:M|=pfor some prime p. So G/M is abelian, and G0 _{≤ M. By the arbitrariness of M we get}

G0 _{≤Φ(G). ¤}

By Wielandt Theorem, if Gis nilpotent then G/Φ(G) is abelain. More- over, we know thatG/M is cyclic of prime order for every maximal subgroup M.

Remark: It is not easy to determine the Frattini subgroup of a given finite group, because for most groups it is hard to determine their all maximal subgroups. In the next section, we shall show that for finite p-groups, it is quite easy to determine their Frattini subgroups. Also for a nonabelian simple group, its Frattini subgroup is trivial, because it does not have any nontrivial normal subgroup; and its derived subgroup is itself. Hence, by Wielandt’s Theorem, it is not nilpotent. Also,A₄ andD_2n, where nis not a power of 2, are not nilpotent. But, if nis a 2-power, then D2n is nilpotent.

(Give a proof!) Exercises

4.3.1. If G/Φ(G) is cyclic, thenGis cyclic.

4.3.2. Let G=D2n=ha, b |an=b2 = 1, bab=a−1i be the dihedral group

and n =pα1

1 · · ·pαss. Show that Φ(G) =hap1p2···psi, which is cyclic of

order pα1−1

1 · · ·pαss−1.

4.3.3. IfN EG, then Φ(G)N/N ≤Φ(G/N). In addition, ifN ≤Φ(G) then Φ(G)/N = Φ(G/N). Give an example to show that ifN 6≤Φ(G), then Φ(G)N/N = Φ(G/N) does not hold.

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