AATA Solvable Groups

Section 10.3 Solvable Groups

Worksheet 10.3.1 Activity: (Sub)Normal Series
A4 US

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

G = H_{n} \supset H_{n - 1} \supset \dots \supset H_{1} \supset H_{0} = {e}

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is a subnormal series provided each

H_{i}

is normal in

H_{i + 1},

and a normal series if each

H_{i}

is normal in

G .

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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.

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1.

Find a subnormal series for $D_{4} .$ Is it a normal series?

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2.

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

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3.

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

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4.

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

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Subsection 10.3.2 Series of Subgroups

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A subnormal series of a group

G

is a finite sequence of subgroups

G = H_{n} \supset H_{n - 1} \supset \dots \supset H_{1} \supset H_{0} = {e},

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where

H_{i}

is a normal subgroup of

H_{i + 1} .

If each subgroup

H_{i}

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.

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Example 10.21.

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

\begin{matrix} Z \supset 9 Z \supset 45 Z \supset 180 Z \supset {0}, \\ Z_{24} \supset ⟨ 2 ⟩ \supset ⟨ 6 ⟩ \supset ⟨ 12 ⟩ \supset {0} . \end{matrix}

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Example 10.22.

A subnormal series need not be a normal series. Consider the following subnormal series of the group $D_{4} :$

D_{4} \supset {(1), (12) (34), (13) (24), (14) (23)} \supset {(1), (12) (34)} \supset {(1)} .

The subgroup ${(1), (12) (34)}$ is not normal in $D_{4};$ consequently, this series is not a normal series.

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A subnormal (normal) series

{K_{j}}

is a refinement of a subnormal (normal) series

{H_{i}}

{H_{i}} \subset {K_{j}} .

That is, each

H_{i}

is one of the

K_{j} .

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Example 10.23.

The series

Z \supset 3 Z \supset 9 Z \supset 45 Z \supset 90 Z \supset 180 Z \supset {0}

is a refinement of the series

Z \supset 9 Z \supset 45 Z \supset 180 Z \supset {0} .

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The best way to study a subnormal or normal series of subgroups,

{H_{i}}

G,

is actually to study the factor groups

H_{i + 1} / H_{i} .

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}} .

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Example 10.24.

The two normal series

\begin{matrix} Z_{60} \supset ⟨ 3 ⟩ \supset ⟨ 15 ⟩ \supset {0} \\ Z_{60} \supset ⟨ 4 ⟩ \supset ⟨ 20 ⟩ \supset {0} \end{matrix}

of the group $Z_{60}$ are isomorphic since

\begin{matrix} Z_{60} / ⟨ 3 ⟩ ≅ ⟨ 20 ⟩ / {0} ≅ Z_{3} \\ ⟨ 3 ⟩ / ⟨ 15 ⟩ ≅ ⟨ 4 ⟩ / ⟨ 20 ⟩ ≅ Z_{5} \\ ⟨ 15 ⟩ / {0} ≅ Z_{60} / ⟨ 4 ⟩ ≅ Z_{4} . \end{matrix}

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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.

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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}}

G

is a principal series if all the factor groups are simple.

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Example 10.25.

The group $Z_{60}$ has a composition series

Z_{60} \supset ⟨ 3 ⟩ \supset ⟨ 15 ⟩ \supset ⟨ 30 ⟩ \supset {0}

with factor groups

\begin{aligned} Z_{60} / ⟨ 3 ⟩ & ≅ Z_{3} \\ ⟨ 3 ⟩ / ⟨ 15 ⟩ & ≅ Z_{5} \\ ⟨ 15 ⟩ / ⟨ 30 ⟩ & ≅ Z_{2} \\ ⟨ 30 ⟩ / {0} & ≅ Z_{2} . \end{aligned}

Since $Z_{60}$ is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series

Z_{60} \supset ⟨ 2 ⟩ \supset ⟨ 4 ⟩ \supset ⟨ 20 ⟩ \supset {0}

is also a composition series.

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Example 10.26.

For $n \geq 5,$ the series

S_{n} \supset A_{n} \supset {(1)}

is a composition series for $S_{n}$ since $S_{n} / A_{n} ≅ Z_{2}$ and $A_{n}$ is simple.

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Example 10.27.

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

{0} = H_{0} \subset H_{1} \subset \dots \subset H_{n - 1} \subset H_{n} = Z

is a subnormal series for the integers under addition. Then $H_{1}$ must be of the form $k Z$ for some $k \in N .$ In this case $H_{1} / H_{0} ≅ k Z$ is an infinite cyclic group with many nontrivial proper normal subgroups.

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Although composition series need not be unique as in the case of

Z_{60},

it turns out that any two composition series are related. The factor groups of the two composition series for

Z_{60}

are

Z_{2},

Z_{2},

Z_{3},

and

Z_{5};

that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

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Theorem 10.28. Jordan-Hölder.

Any two composition series of $G$ are isomorphic.

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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 $1 \leq k < n .$ Let

\begin{matrix} G = H_{n} \supset H_{n - 1} \supset \dots \supset H_{1} \supset H_{0} = {e} \\ G = K_{m} \supset K_{m - 1} \supset \dots \supset K_{1} \supset K_{0} = {e} \end{matrix}

be two composition series for $G .$ 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 $K_{j + 1} \cap H_{n - 1} :$

\begin{matrix} G = H_{n} \supset H_{n - 1} \supset H_{n - 1} \cap K_{m - 1} \supset \dots \supset H_{0} \cap K_{m - 1} = {e} \\ G = K_{m} \supset K_{m - 1} \supset K_{m - 1} \cap H_{n - 1} \supset \dots \supset K_{0} \cap H_{n - 1} = {e} . \end{matrix}

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

\begin{aligned} (H_{i + 1} \cap K_{m - 1}) / (H_{i} \cap K_{m - 1}) & = (H_{i + 1} \cap K_{m - 1}) / (H_{i} \cap (H_{i + 1} \cap K_{m - 1})) \\ ≅ H_{i} (H_{i + 1} \cap K_{m - 1}) / H_{i}, \end{aligned}

where $H_{i}$ is normal in $H_{i} (H_{i + 1} \cap K_{m - 1}) .$ 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 $H_{i} / H_{i} .$ That is, $H_{i} (H_{i + 1} \cap K_{m - 1})$ must be either $H_{i}$ or $H_{i + 1} .$ Removing any nonproper inclusions from the series

H_{n - 1} \supset H_{n - 1} \cap K_{m - 1} \supset \dots \supset H_{0} \cap K_{m - 1} = {e},

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

H_{n - 1} \supset \dots \supset H_{1} \supset H_{0} = {e} .

Hence, the composition series

G = H_{n} \supset H_{n - 1} \supset \dots \supset H_{1} \supset H_{0} = {e}

and

G = H_{n} \supset H_{n - 1} \supset H_{n - 1} \cap K_{m - 1} \supset \dots \supset H_{0} \cap K_{m - 1} = {e}

are equivalent. If $H_{n - 1} = K_{m - 1},$ 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 $H_{n - 1} .$ In this case $H_{n - 1} K_{m - 1} = G$ and we can apply the Second Isomorphism Theorem once again; that is,

K_{m - 1} / (K_{m - 1} \cap H_{n - 1}) ≅ (H_{n - 1} K_{m - 1}) / H_{n - 1} = G / H_{n - 1} .

Therefore,

G = H_{n} \supset H_{n - 1} \supset H_{n - 1} \cap K_{m - 1} \supset \dots \supset H_{0} \cap K_{m - 1} = {e}

and

G = K_{m} \supset K_{m - 1} \supset K_{m - 1} \cap H_{n - 1} \supset \dots \supset K_{0} \cap H_{n - 1} = {e}

are equivalent and the proof of the theorem is complete.

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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.

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Example 10.29.

The group $S_{4}$ is solvable since

S_{4} \supset A_{4} \supset {(1), (12) (34), (13) (24), (14) (23)} \supset {(1)}

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

S_{n} \supset A_{n} \supset {(1)}

is a composition series for $S_{n}$ with a nonabelian factor group. Therefore, $S_{n}$ is not a solvable group for $n \geq 5 .$

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Exercises 10.3.3 Collected Homework

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1.

Consider the normal series below for $Z_{24} :$

Z_{24} \supset ⟨ 12 ⟩ \supset {0} .

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 $G^{'} / ⟨ 12 ⟩$ for some $G^{'}$
Demonstrate/explain how this shows us how to build a longer normal series for $Z_{24} .$
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).

Section 10.3 Solvable Groups

Worksheet 10.3.1 Activity: (Sub)Normal Series A4US

1.

2.

3.

4.

Subsection 10.3.2 Series of Subgroups

Example 10.21.

Example 10.22.

Example 10.23.

Example 10.24.

Example 10.25.

Example 10.26.

Example 10.27.

Theorem 10.28. Jordan-Hölder.

Proof.

Example 10.29.

Exercises 10.3.3 Collected Homework

1.

Worksheet 10.3.1 Activity: (Sub)Normal Series
A4 US