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Section 13.2 Solvable Groups

A subnormal series of a group G is a finite sequence of subgroups
G=HnHn1H1H0={e},
where Hi is a normal subgroup of Hi+1. If each subgroup Hi is normal in G, then the series is called a normal series. The length of a subnormal or normal series is the number of proper inclusions.

Example 13.11.

Any series of subgroups of an abelian group is a normal series. Consider the following series of groups:
Z9Z45Z180Z{0},Z242612{0}.

Example 13.12.

A subnormal series need not be a normal series. Consider the following subnormal series of the group D4:
D4{(1),(12)(34),(13)(24),(14)(23)}{(1),(12)(34)}{(1)}.
The subgroup {(1),(12)(34)} is not normal in D4; consequently, this series is not a normal series.
A subnormal (normal) series {Kj} is a refinement of a subnormal (normal) series {Hi} if {Hi}{Kj}. That is, each Hi is one of the Kj.

Example 13.13.

The series
Z3Z9Z45Z90Z180Z{0}
is a refinement of the series
Z9Z45Z180Z{0}.
The best way to study a subnormal or normal series of subgroups, {Hi} of G, is actually to study the factor groups Hi+1/Hi. We say that two subnormal (normal) series {Hi} and {Kj} of a group G are isomorphic if there is a one-to-one correspondence between the collections of factor groups {Hi+1/Hi} and {Kj+1/Kj}.

Example 13.14.

The two normal series
Z60315{0}Z60420{0}
of the group Z60 are isomorphic since
Z60/320/{0}Z33/154/20Z515/{0}Z60/4Z4.
A subnormal series {Hi} of a group G is a composition series if all the factor groups are simple; that is, if none of the factor groups of the series contains a normal subgroup. A normal series {Hi} of G is a principal series if all the factor groups are simple.

Example 13.15.

The group Z60 has a composition series
Z6031530{0}
with factor groups
Z60/3Z33/15Z515/30Z230/{0}Z2.
Since Z60 is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series
Z602420{0}
is also a composition series.

Example 13.16.

For n5, the series
SnAn{(1)}
is a composition series for Sn since Sn/AnZ2 and An is simple.

Example 13.17.

Not every group has a composition series or a principal series. Suppose that
{0}=H0H1Hn1Hn=Z
is a subnormal series for the integers under addition. Then H1 must be of the form kZ for some kN. In this case H1/H0kZ is an infinite cyclic group with many nontrivial proper normal subgroups.
Although composition series need not be unique as in the case of Z60, it turns out that any two composition series are related. The factor groups of the two composition series for Z60 are Z2, Z2, Z3, and Z5; that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

Proof.

We shall employ mathematical induction on the length of the composition series. If the length of a composition series is 1, then G must be a simple group. In this case any two composition series are isomorphic.
Suppose now that the theorem is true for all groups having a composition series of length k, where 1k<n. Let
G=HnHn1H1H0={e}G=KmKm1K1K0={e}
be two composition series for G. We can form two new subnormal series for G since HiKm1 is normal in Hi+1Km1 and KjHn1 is normal in Kj+1Hn1:
G=HnHn1Hn1Km1H0Km1={e}G=KmKm1Km1Hn1K0Hn1={e}.
Since HiKm1 is normal in Hi+1Km1, the Second Isomorphism Theorem (Theorem 11.12) implies that
(Hi+1Km1)/(HiKm1)=(Hi+1Km1)/(Hi(Hi+1Km1))Hi(Hi+1Km1)/Hi,
where Hi is normal in Hi(Hi+1Km1). Since {Hi} is a composition series, Hi+1/Hi must be simple; consequently, Hi(Hi+1Km1)/Hi is either Hi+1/Hi or Hi/Hi. That is, Hi(Hi+1Km1) must be either Hi or Hi+1. Removing any nonproper inclusions from the series
Hn1Hn1Km1H0Km1={e},
we have a composition series for Hn1. Our induction hypothesis says that this series must be equivalent to the composition series
Hn1H1H0={e}.
Hence, the composition series
G=HnHn1H1H0={e}
and
G=HnHn1Hn1Km1H0Km1={e}
are equivalent. If Hn1=Km1, then the composition series {Hi} and {Kj} are equivalent and we are done; otherwise, Hn1Km1 is a normal subgroup of G properly containing Hn1. In this case Hn1Km1=G and we can apply the Second Isomorphism Theorem once again; that is,
Km1/(Km1Hn1)(Hn1Km1)/Hn1=G/Hn1.
Therefore,
G=HnHn1Hn1Km1H0Km1={e}
and
G=KmKm1Km1Hn1K0Hn1={e}
are equivalent and the proof of the theorem is complete.
A group G is solvable if it has a subnormal series {Hi} such that all of the factor groups Hi+1/Hi are abelian. Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations.

Example 13.19.

The group S4 is solvable since
S4A4{(1),(12)(34),(13)(24),(14)(23)}{(1)}
has abelian factor groups; however, for n5 the series
SnAn{(1)}
is a composition series for Sn with a nonabelian factor group. Therefore, Sn is not a solvable group for n5.