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Chapter 5: Problem 2

Prove that a subgroup of a solvable group is solvable.

Short Answer

Expert verified
A subgroup of a solvable group is also solvable.

Step by step solution

01

Understand the Definition of a Solvable Group

A group \( G \) is considered solvable if it has a subnormal series such that the factor groups (or quotient groups) are all abelian. In simpler terms, this means there exists a sequence of subgroups \( \{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G \) where each factor group \( G_{i+1}/G_i \) is abelian.
02

Consider a Subgroup of a Solvable Group

Let \( H \) be a subgroup of a solvable group \( G \). We need to demonstrate that \( H \) is also solvable. We know that \( G \) has a subnormal series \( \{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G \) with abelian factor groups.
03

Construct a Subnormal Series in the Subgroup

There is a construction of the intersection of each subgroup in the subnormal series of \( G \) with \( H \). Consider the series \( \{1\} = H_0 \subseteq H_1 \subseteq \cdots \subseteq H_n = H \) where \( H_i = H \cap G_i \).
04

Check Each Factor in the Intersection Series

Because each \( G_{i+1}/G_i \) is abelian and affects \( H \), the factor groups \( H_{i+1}/H_i = (H \cap G_{i+1})/(H \cap G_i) \) are subgroups of \( G_{i+1}/G_i \). Hence they are also abelian.
05

Conclude that the Subgroup is Solvable

Since there is a series of subgroups \( \{1\} = H_0 \subseteq H_1 \subseteq \cdots \subseteq H_n = H \) with all factor groups being abelian, \( H \) is solvable by definition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subgroup
In group theory, a subgroup is simply a smaller group within a larger group. It's formed from the same elements adhering to the same operation as the parent group, but with fewer elements. Think of a subgroup as a miniature version of its parent group. To qualify as a subgroup, two main conditions must be satisfied:
  • Closure: If you take any two elements in the subgroup and perform the group operation, you'll end up with another element within the same subgroup.
  • Contains the Identity: A subgroup must always have the identity element of the parent group, which is the element that leaves other elements unchanged when used in an operation.
Subgroups are crucial because they help in analyzing the structure of more complex groups by breaking them down into simpler, more manageable components. In the context of solvable groups, any subgroup of a solvable group will also exhibit solvability. This is due to the inherited structure from the parent group.
Subnormal Series
A subnormal series is like a roadmap that helps us navigate through the structure of a group, showcasing its layered nature. For a group to be considered solvable, it must possess a subnormal series with certain properties. In this series,
  • Each subgroup is a normal subgroup of the subsequent group, meaning it fits nicely within the larger group structure.
  • The sequence typically starts from the trivial group \( \{1\} \) and moves through a progression of subgroups until it reaches the entire group \( G \).
So, a subnormal series from \( G \) looks like: \( \{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G \). The presence of such a series in a group indicates that the group can be "built up" in a very orderly manner. For a subgroup derived from a solvable group, we can construct a comparable subnormal series that shows it is also solvable.
Abelian Group
The term "abelian group" refers to a special kind of group where the order of performing operations doesn't change the outcome. Here, for any two elements \( a \) and \( b \) in the group, the equation \( a \cdot b = b \cdot a \) holds. This property, known as commutativity, makes computations within abelian groups straightforward.
  • Think of it like addition with real numbers where \( 3 + 5 \) is the same as \( 5 + 3 \).
  • Abelian groups are often used as building blocks in theoretical aspects of mathematics and group theory.
In the context of solvable groups, the key is that the factor groups derived from a subnormal series must all be abelian. This property simplifies the structure of the larger group, making it more manageable and predictable.
Factor Group
A factor group, or quotient group, emerges from the process of dividing a group by one of its normal subgroups. In simpler terms, it's like compressing a group by merging elements into sets based on an equivalence relation derived from the subgroup. For a group \( G \) and a normal subgroup \( N \), the factor group is written as \( G/N \).
  • The elements of \( G/N \) are these merged sets, formally called cosets.
  • This construction allows us to analyze the group鈥檚 structure in a compressed form, focusing on the interaction between different cosets rather than individual elements.
In solvable groups, each factor group within the subnormal series is required to be an abelian group. This ensures a step-by-step simplification of the group structure, aligning with the concept of solvability by reducing complexity through each transition in the series.

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Most popular questions from this chapter

If \(F\) is of characteristic 0 and \(f(x) \in F[x]\) is such that \(f^{\prime}(x)=0\), prove that \(f(x)=\alpha \in F\)

Prove that a line in \(F\) and a circle in \(F\) which intersect in the real plane do so at a point either in the plane of \(F\) or in the plane of \(F(\sqrt{\gamma})\) where \(\gamma\) is a positive number in \(F\).

If \(F\) is the field of rational numbers, find necessary and sufficient conditions on \(a\) and \(b\) so that the splitting field of \(x^{3}+a x+b\) has degree exactly 3 over \(F\).

(a) Let \(R\) be the feld of real numbers and \(Q\) the field of rational numbers. In \(R, \sqrt{2}\) and \(\sqrt{3}\) are both algebraio over \(Q .\) Exhibit a polynomial of degree 4 over \(Q\) satisfied by \(\sqrt{2}+\sqrt{3}\). (b) What is the degree of \(\sqrt{2}+\sqrt{3}\) over \(Q ?\) Prove your answer. (c) What is the degree of \(\sqrt{2} \sqrt{3}\) over \(Q ?\)

Let \(p\) be a prime number and let \(F=J_{p}\), the field of integers mod \(p\). (a) Prove that there is an irreducible polynomial of degree 2 over \(F\). (b) Use this polynomial to construct a field with \(p^{2}\) elements. (c) Prove that any two irreducible polynomials of degree 2 over \(F\) lead to isomorphic fields with \(p^{2}\) elements.
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