Problem 18
Question
Let \(H\) be a subgroup of a group \(G\) and let \(X\) be the set \(\\{x H \mid x \in G\\}\) of all left cosets of \(H\) in \(G\). Let \(G\) act on \(H\) by left multiplication \((g, x H) \rightarrow g x H \in X\). Let \(i=[G: H]\) be the index of \(H\) in \(G\). Then (a) Show that if \(\chi\) is one to one, then \(|G|\) divides \(i !\) (b) Show that if \(|G|\) does not divide \(i !\), then the kernel \(K\) is nontrivial. (c) Show that if \(|G|\) does not divide \(i !,\) then \(G\) has a nontrivial proper normal subgroup.
Step-by-Step Solution
Verified Answer
Part (a): |G| divides i!. Part (b): K is nontrivial. Part (c): G has a nontrivial normal subgroup.
1Step 1: Understand the mapping χ
Define the function \(\chi: G \to \text{Sym}(X)\), mapping each element of \(G\) to a permutation of left cosets. Here, \(\text{Sym}(X)\) is the symmetric group on \(|X|\) elements. If \(\chi\) is one-to-one, \(|G|\) must equal the order of \(\chi(G)\), which is a permutation subgroup of \(\text{Sym}(X)\).
2Step 2: The importance of i = [G : H]
Recall that \(i = [G : H]\) is the number of left cosets of \(H\) in \(G\), which equals \(|X|\). The symmetric group \(\text{Sym}(X)\) has order \(i!\).
3Step 3: Proving Part (a)
If \(\chi\) is injective, \(G\) is isomorphic to a subgroup of \(\text{Sym}(X)\). Therefore, \(|G|\), being a divisor of the order of its image, must divide \(|\text{Sym}(X)| = i!\).
4Step 4: Proving Part (b)
The kernel of \(\chi\), \(K\), contains all elements of \(G\) that act trivially on \(X\). If \(|G|\) does not divide \(i!\), then \(\chi\) cannot be injective; thus, \(K\) must be nontrivial, forming a subgroup of \(G\).
5Step 5: Proving Part (c)
If the kernel \(K\) is nontrivial and normal (since \(\chi\) is a homomorphism), \(K\) is a nontrivial proper normal subgroup of \(G\). A group homomorphism with a nontrivial kernel always indicates the presence of a nontrivial normal subgroup.
Key Concepts
SubgroupCosetsSymmetric GroupGroup Action
Subgroup
In group theory, a **subgroup** is a smaller group within a larger group that maintains the properties of the larger group. It follows the same operation rules as its parent group. If we have a group \(G\) and a subset \(H\) of \(G\), \(H\) is considered a subgroup if it satisfies the properties:
Subgroups are often used to divide a group into smaller, more manageable pieces, which can simplify many aspects of group theory analysis.
- Closure: For any two elements \(a, b\) in \(H\), the product \(ab\) is also in \(H\).
- Identity: The identity element of \(G\) is also in \(H\).
- Inverse: For every element \(a\) in \(H\), its inverse \(a^{-1}\) is also in \(H\).
Subgroups are often used to divide a group into smaller, more manageable pieces, which can simplify many aspects of group theory analysis.
Cosets
Cosets are a fundamental concept in group theory, especially when dealing with subgroups. If you have a group \(G\) and a subgroup \(H\), you can form **left cosets** as follows: for any element \(x\) in \(G\), the left coset of \(H\) containing \(x\) is the set \(xH = \{xh \, | \, h \in H\}\).
Cosets are important because they help partition the entire group \(G\) into non-overlapping subsets of equal size, facilitating more straightforward calculations and insights into the group's structure:
Cosets are important because they help partition the entire group \(G\) into non-overlapping subsets of equal size, facilitating more straightforward calculations and insights into the group's structure:
- Each coset has the same number of elements as the subgroup \(H\).
- Cosets can help determine whether a subgroup is a normal subgroup, which influences how groups can be broken down into simpler components.
- Understanding cosets leads to concepts like group actions and provides tools for counting arguments in group theory.
Symmetric Group
The **symmetric group**, denoted \(\text{Sym}(X)\), on a set \(X\) is the group of all permutations of \(X\). A permutation is a bijective function from the set to itself. In simpler terms, it's a way of rearranging elements of a set.
The symmetric group has several key properties that make it important in group theory:
The symmetric group has several key properties that make it important in group theory:
- The number of elements (order) of \(\text{Sym}(X)\) is the factorial of the number of elements in \(X\), represented as \(|X|!\).
- Symmetric groups are special because any group can be made to behave like a subgroup of some symmetric group due to Cayley's theorem.
- Symmetric groups are also central in understanding group actions, as they provide examples of how elements can act on sets.
Group Action
A **group action** occurs when a group \(G\) acts on a set \(X\) and rearranges its elements according to a group operation. It is a formal way to express the idea of symmetry and transformation.
For a group \(G\) to act on a set \(X\), there needs to be a rule (action) such that:
They can provide information about the structure of the group itself and lead to the discovery of properties like invariants and orbits, crucial concepts in areas like geometry, physics, and chemistry.
For a group \(G\) to act on a set \(X\), there needs to be a rule (action) such that:
- For every element \(g\) in \(G\) and \(x\) in \(X\), there is an element \(g \cdot x\) in \(X\).
- The action respects the group operation: \((g_1g_2) \cdot x = g_1 \cdot (g_2 \cdot x)\).
- The identity element of \(G\) acts as the identity on \(X\): \(e \cdot x = x\) for all \(x\) in \(X\).
They can provide information about the structure of the group itself and lead to the discovery of properties like invariants and orbits, crucial concepts in areas like geometry, physics, and chemistry.
Other exercises in this chapter
Problem 17
Pertain to Example \(4.7 .16,\) were we have a group \(G\) of order 30, an element \(a\) of order \(2,\) an element \(b\) of order \(15,\) and a relation of for
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Show that no group \(G\) of order \(|G|=n\) is simple where (a) \(n=45\) (b) \(n=16\) (c) \(n=p^{r}, p\) prime, \(r>1\) (d) \(n=p^{r} m, p\) prime, \(r \geq 1,
View solution Problem 18
Let \(r\) be the index \([G: Z(G)]\) of the center \(Z(G)\). Show that for any \(g \in G\), the number of elements \(|K(g)|\) of the conjugacy class \(K(g)\) of
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Explain why for any finite group \(G\) and any \(g \in G,\) the number of elements \(|K(g)|\) of the conjugacy class \(K(g)\) of the element \(g\) divides the o
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