Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>. Assume that the result is not true, and let G be a counterexample of minimal order. Since Gi (i=1,2,3) is supersolvable, it follows that Gi has a normal Sylow p-subgroup, where p is the largest prime dividing |G|. Then, by Lemma 2.1, G has a normal Sylow p-subgroup, say P. Certainly, every proper quotient group of G satisfies the hypothesis of the theorem. So every proper quotient group of G is supersolvable by the minimal choice of G. But the class of all supersolvable groups is a saturated formation, so Φ(G)=1, P=F(G), and CG(P)=P. We argue that G1∩G2≠1. If not, G1∩G2=1. Then, G=G1G2 is a totally permutable product of G1 and G2 by Lemma 2.2(a). Then, by Lemma 2.3, G is supersolvable, a contradiction.

Thus, G1∩G2≠1. Analogously, G1∩G3≠1 and G2∩G3≠1. Since G=G1G2 is a mutually permutable product of G1 and G2, it follows that G1∩G2 is a quasinormal subgroup of G1 and of G2 by Lemma 2.2(b). Then, by Lemma 2.4, G1∩G2 is a subnormal subgroup of G1 and of G2. Hence G1∩G2 is a subnormal subgroup of G by Lemma 2.5.

If CorG1 (G1∩G2)=1, then, by Lemma 2.6, G1∩G2 is nilpotent. Hence, G1∩G2 is a subnormal nilpotent subgroup of G. So G1∩G2≤P by Lemma 2.7. Taken into consideration that G1∩G2 is quasinormal in G1, G1∩G2≤P and P is abelian, it follows that G1∩G2 is normal in G1 and so G1∩G2=CorG1 (G1∩G2)=1, a contradiction. Thus CorG1 (G1∩G2)≠1. Analogously, CorG2 (G2∩G3)≠1 and CorG3 (G3∩G1)≠1. Set L1=CorG1 (G1∩G2), L2=CorG2 (G2∩G3) and L3=CorG3 (G3∩G1). Then L1G=L1G1G2≤G2, L2G=L2G2G3≤G3, and L3G=L3G3G1≤G1. Now, P≤LiG (i=1,2,3) as P is a unique minimal normal subgroup of G. Hence, P≤Gi (i=1,2,3).

Now, we finish the proof of the theorem. Since Gi is supersolvable, CG(P)=P=F(G) and P≤Gi (i=1,2,3), it follows that CGi(P)=CGi(F(Gi))=P (i=1,2,3) and Gi/P is abelian. Hence, by Lemma 2.8, G/P is nilpotent, and, since the Sylow subgroups of G are abelian, it follows that G/P is abelian. On the other hand, by Lemma 2.9, Gi/P (i=1,2) is of exponent dividing p-1. Hence, G/P is abelian of exponent dividing p-1 and so G is supersolvable, by Lemma 2.10, a final contradiction completing the proof of the theorem.

Proof of Corollary <xref ref-type="statement" rid="coro1.2">1.2</xref>. Assume that the result is not true and let G be a counterexample of minimal order. Let P be a Sylow p-subgroup of G, where p is the largest prime dividing |G|. Then, P is normal in G by Lemma 2.1. Certainly, every proper quotient group of G satisfies the hypothesis of the corollary. So every proper quotient group of G is supersolvable by the minimal choice of G. But the class of all supersolvable groups is a saturated formation, so Φ(G)=1, P=F(G), and CG(P)=P. Since G1, G2, and G3 have coprime indices, we can assume that p does not divide |G:G1| and p does not divide |G:G2|. Then, P≤G1 and P≤G2 and so P=CG1(P)=CG2(P)=F(G2)=F(G1) as P=F(G)=CG(P). Then, G1/P and G2/P are abelian subgroups of G/P by Lemma 2.11. This together with (|G:G1|,|G:G2|)=1 imply that the Sylow subgroups of G/P are abelian. Now Theorem 1.1 implies that G is supersolvable, a contradiction completing the proof of the corollary.