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

 (11.7)

be a sequence of subgroups of the group where each is normal in (but not necessarily in all of ). Such a sequence of subgroups is called a normal series for . Associated with a normal series for a group , is an associated sequence of factors (or factor groups); viz

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:

 (11.8)

where is the Klein -group (see Section 6.3), and, as usual, denotes a cyclic group of order , here, e.g., take . Note that all terms in the fifth series given above for occur in the fourth and all those in the fourth occur in the third and those in the third occur in the first. A similar situation prevails between the fifth, fourth, second, and first series. This illustrates the following: one normal series is called a refinement of another if all the terms of the second occur in the first series. Hence the second series above is a refinement of the fourth series. The third series is also a refinement of the fourth series. However, the second series is not a refinement of the third series.

Finally, two normal series

 (11.9)

are called equivalent or (isomorphic) if there exists a 1-1 correspondence between the factors of the two series (thus ) such that the corresponding factors are isomorphic.

Example 11.2.1   Consider the two series for ,

where denotes the residue class of mod . These normal series are equivalent: For , , while the factors of the second series are , (where denotes the residue class of mod ).

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

Theorem 11.2.2 (Schreier)   Any two normal series for a group G have equivalent refinements.

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.

Theorem 11.2.3   If is a normal series for the group and if , then

is a normal series for . The factors of the normal series in (11.2.3) are isomorphic to subgroups of the factors of the normal series for .

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).

Definition 11.2.4   A group is said to be solvable if it has a normal series all of whose factors are abelian groups.

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.

Theorem 11.2.5   Any subgroup and any factor group of a solvable group is solvable.

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
 (11.10)

By Schreier's Theorem (Theroem 11.2.2), (11.7) and (11.10) have equivalent refinements. Let
 (11.11)

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.

Subsections

David Joyner 2007-08-06