Let be an -module. A finite sequence of submodules
is called normal series of .
Note that a normal series of a module is a normal series of the underlying group ; indeed, each subgroup of an abelian group is normal, hence each normal series in modules gives rise to a normal series of groups. The other direction is not true, since additive subgroups need not be closed under multiplication by elements of .
Let be an -module, and let a normal series
be given. This normal series is said to be without repetitions if each inclusion of modules is, in fact, strict.
Refinements and composition series
Let be an -module, and let a normal series
be given. A refinement of this normal series is another normal series
such that .
Note that this implies . Refinements arise from a normal series
by inserting submodules between two modules of the composition series and ; that is, we start with two modules of a composition series and , find a submodule of such that , and then just insert this into the normal series.
Let be an -module. We say that is simple if and only if it has no proper submodules (i.e. no submodules of which are neither equal to nor equal to ).
Let be an -module. A composition series of is a normal series of , say
- there are no repetitions, and
- each so-called composition factor of that series, namely for , is simple (where we set and ).
Equivalently, a composition series is a normal series without repetitions, such that any proper refinement of it has repetitions.
To any module, we may associate a so-called length. This concept is justified by the following theorem:
Theorem 12.6 (Jordan):
Let be an -module which has a composition series
We say that this composition series has length , and then it follows that
- each normal series in without repetitions has length ,
- every other composition series of also has length , and
- every normal series in can be refined to a composition series of .
First, we note that 1. implies 3., since whenever a normal series has a refinement that has no repetitions, we may apply that refinement, and due to 1., we must eventually reach a composition series.
Then we prove 1. and 2. by induction on . Indeed, for , this theorem follows since then is simple, and therefore any normal series of length must have repetitions, which is why the trivial normal series is the only one without repetitions, and there is only one composition series.
Assume now the case to be valid. Let there be a composition series
of length , and assume that there is any other normal series
without repetition of length . Now hence has a composition series of length . By induction, we have:
- If , then is a normal series in and hence has length at most , whence the complete normal series has length at most .
- If , then has a normal series without repetitions of length , which is a contradiction.
- If not , we have , for otherwise the composition series would have a proper refinement. Then we have two normal series
- Now has a composition series of length , whence has a composition series of length . Furthermore, , which is why any such composition series then extends to a composition series of of length . Therefore, the partial series
- has length at most .
This proves 1. by induction. Furthermore, by induction, can not have a composition series of length , since then also the composition series above would have length , whence 2. is proven by 1. and induction.
Assume has a composition series. Then the length of the module is defined to be the length of such a composition series.
If doesn't have a composition series, we set the length of to be .
Furthermore, composition series are essentially unique, as given by the following theorem:
Theorem 12.8 (Hölder):
are two composition series, then there exists a permutation such that for all
(again and , and analogous for ).
We say that the two series are equivalent.
We proceed by induction on . For , we have only the trivial composition series as composition series. Now assume the theorem for . Let two composition series
be given. If , we have equivalence by induction. If not, we have once again (since neither can be properly contained in the other, for else we would obtain a contradiction to the previous theorem of Jordan). Now must have a composition series, since by the previous theorem we may refine the series
to a composition series of . Further, we again have
- and ;
both modules on the right of the isomorphisms are simple, whence we get two composition series of given by
Now the two above isomorphisms also imply that these two are equivalent, and by induction, the first one is equivalent to the first composition series, and the second one equivalent to the second composition series.
Let be an -module, let be a submodule.
- has a composition series if and only if both and have composition series.
If has a composition series, then intersecting this series or projecting this series gives normal series of or respectively. When the repetitions are crossed out, no refinements are possible (else they induce a refinement of the original composition series, in the latter case by the correspondence theorem).
If and both have composition series, we take a composition series
of and another one of given by
By the correspondence theorem, we write for suitable . Then
is a composition series of .
Normal series between modules
Let be a module and a submodule. A series
is called a normal series between and , and if each inclusion is strict and there does not exist a refinement which leaves each inclusion strict, it is called a composition series between and .
By the correspondence theorem, we get a bijection between normal (or composition) series
between and on the one hand, and of normal (or composition) series
of . Then by the above and the third isomorphism theorem, composition series between and are essentially unique. Further, if there is a composition series, normal series can be refined to composition series of same length.