Let be a finite group. Prove that the following are equivalent.

- is solvable.
- has a subnormal series such that is cyclic for .
- All composition factors in any composition series of have prime order.
- has a subnormal series such that is normal in and is abelian for .

We begin with some lemmas.

Lemma 1: Let be a finite group and a normal subgroup such that is abelian. Then there exists a chain of subgroups such that has prime order for all . Proof: is a finite group, so by a previous exercise there exists a composition series such that is simple for all . Now because is abelian, each is abelian. We saw in a previous exercise that the only abelian simple groups are (up to isomorphism) for primes , each has prime order. Using the Lattice Isomorphism Theorem we have a chain of subgroups and for all , by the Third Isomorphism Theorem.

Lemma 2: If is simple and solvable, then for some prime . Proof: If is simple, then the only subnormal series of is . Then is abelian; we saw previously that every abelian simple group is isomorphic to for some prime .

Lemma 3: If is a composition series where is solvable and finite, then is abelian for all . Proof: We proceed by induction on . If , then and the conclusion that is abelian for all holds vacuously. Suppose now that for all composition series of length such that is solvable we have all composition factors abelian, and let be a composition series such that is solvable. Now is solvable by a previous exercise, and is a composition series for of length . By the induction hypothesis, we have abelian for all . Moreover, is simple and solvable by a previous exercise, so that is abelian. By induction the conclusion holds for all composition series which terminate with a solvable group.

Lemma 4: Let be a group and a subgroup. Then is a normal subgroup of . Proof: Suppose . Then for all , we have and for some . Now ; by the Subgroup criterion, is a subgroup of . Now let ; then ; hence this subgroup is normal in .

Lemma 5: Let be a finite nonsimple group such that every composition factor of every composition series of has prime order and let be a minimal nontrivial normal subgroup. Then is abelian. Proof: Let be such a subgroup of . By a previous exercise, there exists a composition series of which contains as a term, say . Let ; by our hypothesis, has prime order and thus is isomorphic to the abelian simple group . Thus for all , we have , hence . Now let and be arbitrary. Because is normal in , we have and . By the previous point, . Thus , hence . Now by Lemma 4, is normal in . Because is a minimal nontrivial normal subgroup of and is strictly contained in , we have , hence , hence . Thus is abelian.

Now to the main result. We prove the following chains of implications: and .

Note that because every group of prime order is cyclic and because every cyclic group is abelian. is trivial.

Let be solvable and let be an arbitrary composition series of . By Lemma 3, all composition factors are abelian simple groups. We saw in a previous exercise that every such group is isomorphic to for some prime , thus all composition factors have prime order.

Let be a finite group with the property that every composition factor of every composition series has prime order.

If , then and it is vacuously true that has a normal series with all intermediate quotients abelian.

Suppose that every finite group with has a normal series with all intermediate quotients abelian, and supose is a group of order .

If is simple, then has only one composition series, namely , and so has prime order by the hypothesis. Thus for some prime and is abelian; so a normal series with all intermediate quotients abelian exists.

If is not simple, then the set of nontrivial normal subgroups of is not empty. Now contains a minimal element which, by Lemma 5, is abelian. By the induction hypothesis, since is a finite group of order less than or equal to , we have a normal series where is abelian. By the Third Isomorphism Theorem, is abelian, and by the Lattice Isomorphism Theorem, is normal. Thus is a normal series for with all intermediate quotients abelian.

## Comments

Another approach to 1=>3: Let be solvable. It has a subnormal series with abelian factor groups. By Schreier refinement theorem, it has a refinement to a composition series, The factor groups of this series are abelian since they are quotients of abelian groups. So these are simple abelian, thus have prime orders by a previous exercise(3.4#1).

very nice