Series of groups; solvable groups revisited

We shall apply the third isomorphism theorem presently, but first we introduce a special sequence of subgroups of a group (usually called **series** of groups). This was alluded to in the introduction to this chapter. Let

We observe that a normal series always exists for an arbitrary group . We could, e.g., take the trivial normal series: . There is also nothing unique about a normal series, e.g., the symmetric group has the following normal series, among others:

Finally, two normal series

Our first general theorem in this spirit is the following important theorem due to Schreier.

**Proof:** Consider two normal series for G as in (11.9). Define

Then we have

and

Now applying the third isomorphism theorem (Theorem 11.1.1) to the groups , , , , we have that and . Furthermore, also by Theorem 11.1.1,

Thus the above two are normal series which are refinements of the two given series and they are equivalent.

We shall, in the next section, apply Schreier's Theorem 11.2.2 to obtain the important theorem of Jordan-Hölder, but first we wish to give an alternate characterization of solvable groups. Before doing this, we establish the following useful theorem.

**Proof:** We apply the second isomorphism theorem (Theorem 8.3.7) to the subgroups and of the group . Hence, we have

First note that since , is a subgroup of , Proposition 8.3.6 implies , and also . Second note that since and that is how we get the group at the lower vertex of our diagram. Thus Theorem 8.3.7 yields that

We now make the following definition (cf. Definition 8.1.6).

Since we have already defined a solvable group in Section 8.1, we must show that these two definitions are equivalent. Thus suppose that is solvable according to Definition 8.1.6. Then

where the superscripts designate the higher commutator subgroups (see Section 8.1). This is, of course, a normal series (see comments at the beginning of Section 8.1 and Theorem 8.1.5) for . Moreover, is abelian by Theorem 8.1.4. Hence is solvable according to Definition 11.2.4 above.

Suppose now that G is solvable in the sense of Defintion 11.2.4. So that G has a normal series as in (11.7), such that each factor is abelian. In particular, is abelian. Thus by Theorem 8.1.4, . Since is abelian, we have, again by Theorem 8.1.4, that . Similarly,

Continuing in this fashion, we finally get that

Hence , and is solvable according to our original definition.

Thus we are at liberty to use whichever characterization of solvabilty is more convenient. In the following theorem, we arbitrarily use the characterization of solvability introduced in this section. We strongly advise the reader to prove the theorem (see exercise 3 for this section) using the initial definition (Definition 8.1.6) without making use of the equivalent characterization we have just established.

**Proof:** Suppose is solvable. Then has a normal series (11.7), such that is abelian. Let . Then form the series (11.2.3). By Theorem 11.2.3, we thus get a normal series, and is isomorphic to a subgroup of and is, therefore, abelian. This completes the first part of the theorem.

Again let be a solvable group and let (11.7) again denote a normal series for with abelian factors. It is easy to see that any refinement of the series (11.7) also has abelian factors; e.g., suppose

is a refinement of (11.7). Then and, hence, is abelian. Since , by Corollary 8.3.5, therefore, also abelian. Now let . Consider the normal series

By Schreier's Theorem (Theroem 11.2.2), (11.7) and (11.10) have equivalent refinements. Let

be a refinement of (11.10) equivalent to a refinement of (11.7). By our preceding observations, the factors of (11.11) are abelian. Since , by Corollary 8.3.5,

is a normal series for with abelian factors.

Since for was shown to be a simple group (see Theorem 6.3.2),

is the only normal series for , when . But for is, of course, non-abelian, hence is not a solvable group. Consequently, by the preceding Theorem 11.2.5, for is also not a solvable group.

David Joyner 2007-08-06