## Section 10.3 Solvable Groups

### Worksheet 10.3.1 Activity: (Sub)Normal Series

Given a group, we can look at subgroups. We say that a sequence of subgroups

is a subnormal series provided each \(H_i\) is normal in \(H_{i+1}\text{,}\) and a normal series if each \(H_i\) is normal in \(G\text{.}\)

A non-trivial group \(G\) is called simple provided it has no non-trivial normal subgroups. We say that a subnormal series is a composition series and that a normal series is a principle series if every quotient group \(H_{i+1}/H_i\) is simple.

###### 1.

Find a subnormal series for \(D_4\text{.}\) Is it a normal series?

###### 2.

Find two different normal series for \(\Z_{60}\) of length 3 (length is the number of proper inclusions).

###### 3.

Find the quotient groups \(H_{i+1}/H_i\) for both series above. How are these related? Are the series composition series?

###### 4.

Find a composition series for \(\Z_{60}\text{.}\) Can you take it to be a refinement of the normal series you found above?

### Subsection 10.3.2 Series of Subgroups

A subnormal series of a group \(G\) is a finite sequence of subgroups

where \(H_i\) is a normal subgroup of \(H_{i+1}\text{.}\) If each subgroup \(H_i\) is normal in \(G\text{,}\) then the series is called a normal series. The length of a subnormal or normal series is the number of proper inclusions.

###### Example 10.21.

Any series of subgroups of an abelian group is a normal series. Consider the following series of groups:

###### Example 10.22.

A subnormal series need not be a normal series. Consider the following subnormal series of the group \(D_4\text{:}\)

The subgroup \(\{ (1), (12)(34) \}\) is not normal in \(D_4\text{;}\) consequently, this series is not a normal series.

A subnormal (normal) series \(\{ K_j \}\) is a refinement of a subnormal (normal) series \(\{ H_i \}\) if \(\{ H_i \} \subset \{ K_j \}\text{.}\) That is, each \(H_i\) is one of the \(K_j\text{.}\)

###### Example 10.23.

The series

is a refinement of the series

The best way to study a subnormal or normal series of subgroups, \(\{ H_i \}\) of \(G\text{,}\) is actually to study the factor groups \(H_{i+1}/H_i\text{.}\) We say that two subnormal (normal) series \(\{H_i \}\) and \(\{ K_j \}\) of a group \(G\) are isomorphic if there is a one-to-one correspondence between the collections of factor groups \(\{H_{i+1}/H_i \}\) and \(\{ K_{j+1}/ K_j \}\text{.}\)

###### Example 10.24.

The two normal series

of the group \({\mathbb Z}_{60}\) are isomorphic since

A group is called simple provided it contains no non-trivial normal subgroups. For series, we care whether the factor groups are simple or not.

A subnormal series \(\{ H_i \}\) 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 \(\{ H_i \}\) of \(G\) is a principal series if all the factor groups are simple.

###### Example 10.25.

The group \({\mathbb Z}_{60}\) has a composition series

with factor groups

Since \({\mathbb Z}_{60}\) is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series

is also a composition series.

###### Example 10.26.

For \(n \geq 5\text{,}\) the series

is a composition series for \(S_n\) since \(S_n / A_n \cong {\mathbb Z}_2\) and \(A_n\) is simple.

###### Example 10.27.

Not every group has a composition series or a principal series. Suppose that

is a subnormal series for the integers under addition. Then \(H_1\) must be of the form \(k {\mathbb Z}\) for some \(k \in {\mathbb N}\text{.}\) In this case \(H_1 / H_0 \cong k {\mathbb Z}\) is an infinite cyclic group with many nontrivial proper normal subgroups.

Although composition series need not be unique as in the case of \({\mathbb Z}_{60}\text{,}\) it turns out that any two composition series are related. The factor groups of the two composition series for \({\mathbb Z}_{60}\) are \({\mathbb Z}_2\text{,}\) \({\mathbb Z}_2\text{,}\) \({\mathbb Z}_3\text{,}\) and \({\mathbb Z}_5\text{;}\) that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

###### Theorem 10.28. Jordan-Hölder.

Any two composition series of \(G\) are isomorphic.

###### 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 \(k\text{,}\) where \(1 \leq k \lt n\text{.}\) Let

be two composition series for \(G\text{.}\) We can form two new subnormal series for \(G\) since \(H_i \cap K_{m-1}\) is normal in \(H_{i+1} \cap K_{m-1}\) and \(K_j \cap H_{n-1}\) is normal in \(K_{j+1} \cap H_{n-1}\text{:}\)

Since \(H_i \cap K_{m-1}\) is normal in \(H_{i+1} \cap K_{m-1}\text{,}\) the Second Isomorphism Theorem implies that

where \(H_i\) is normal in \(H_i (H_{i+1} \cap K_{m-1})\text{.}\) Since \(\{ H_i \}\) is a composition series, \(H_{i+1} / H_i\) must be simple; consequently, \(H_i (H_{i+1} \cap K_{m-1})/ H_i\) is either \(H_{i+1}/H_i\) or \(H_i/H_i\text{.}\) That is, \(H_i (H_{i+1} \cap K_{m-1})\) must be either \(H_i\) or \(H_{i+1}\text{.}\) Removing any nonproper inclusions from the series

we have a composition series for \(H_{n-1}\text{.}\) Our induction hypothesis says that this series must be equivalent to the composition series

Hence, the composition series

and

are equivalent. If \(H_{n-1} = K_{m-1}\text{,}\) then the composition series \(\{H_i \}\) and \(\{ K_j \}\) are equivalent and we are done; otherwise, \(H_{n-1} K_{m-1}\) is a normal subgroup of \(G\) properly containing \(H_{n-1}\text{.}\) In this case \(H_{n-1} K_{m-1} = G\) and we can apply the Second Isomorphism Theorem once again; that is,

Therefore,

and

are equivalent and the proof of the theorem is complete.

A group \(G\) is solvable if it has a subnormal series \(\{ H_i \}\) such that all of the factor groups \(H_{i+1} / H_i\) are abelian. Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations.

###### Example 10.29.

The group \(S_4\) is solvable since

has abelian factor groups; however, for \(n \geq 5\) the series

is a composition series for \(S_n\) with a nonabelian factor group. Therefore, \(S_n\) is not a solvable group for \(n \geq 5\text{.}\)

### Exercises 10.3.3 Collected Homework

###### 1.

Consider the normal series below for \(\Z_{24}\text{:}\)

Find the two quotient groups for the series. Find the “standard” abelian groups each is isomorphic to.

For the

*quotient group*that is not simple found above, find a non-trivial normal subgroup (of the quotient group). Then realize the subgroup as a quotient group \(G'/\langle 12\rangle\) for some \(G'\)Demonstrate/explain how this shows us how to build a longer normal series for \(\Z_{24}\text{.}\)

Find two different composition series for \(\Z_{24}\) (one can be an extension of what you were working on above). Then use quotient groups to demonstrate that the two series are “isomorphic” (and explain what this means).