Problem 6 - A Book of Abstract Algebra

Question

Let \(G\) denote a group, and \(H\) a subgroup of \(G\). Prove the following: Let \(S\) be the union of all the cosets \(\mathrm{Ha}\) such that \(H a=a H\). Then \(S\) is a normal subgroup of \(G\).

Step-by-Step Solution

Answer

The set \( S \) is a normal subgroup of \( G \).

Step 1: Understanding Normal Subgroups

A subgroup \( S \) of \( G \) is called normal if for every element \( g \) of \( G \) and for every element \( s \) of \( S \), the element \( gsg^{-1} \) is also in \( S \). This means that the left cosets and right cosets of \( S \) in \( G \) are identical.

Step 2: Define the Set \(S\)

The set \( S \) is defined as the union of all cosets \( Ha \) of \( H \) for which \( Ha = aH \). This means \( S = \bigcup \{ Ha \mid Ha = aH \} \). Since these cosets are identical from both sides, we know that each coset \( Ha \) is a group in itself.

Step 3: Verify Closure under Group Operation

To show \( S \) is a subgroup, we must show closure under group operation. Take any \( x, y \in S \). Then \( x \in Ha \) and \( y \in Hb \) for such cosets. Their product \( xy \in Hab = H(ab) \). Since \( (ab)H = H(ab) \) due to the nature of the union set \( S \), \( xy \) is in \( S \).

Step 4: Check for Inverses

For any element \( s \in S \), if \( s \in Ha \), then \( s^{-1} \) must also be in \( a^{-1}H = Ha^{-1} = a^{-1}H \) because \( Ha = aH \) implies \( H^{-1}a^{-1} = a^{-1}H\). Thus, \( s^{-1} \in S \).

Step 5: Show Normality

To show \( S \) is normal, take any \( g \in G \) and \( s \in S \). We need to show \( gsg^{-1} \in S \). Since \( s \in Ha \), \( gsg^{-1} \in gHa g^{-1} = g(ah)g^{-1} = gahg^{-1} \). If \( Ha = aH \), we have \( gabg^{-1}a^{-1} = e \) shows the cosets don't affect \( H \). Hence \( Ha \) remains invariant.

Key Concepts

Group Theory

Group Theory is the study of algebraic structures known as groups. A group, denoted by a set, is combined with an operation that satisfies four essential properties: closure, associativity, identity, and invertibility. To have a comprehensive understanding of groups:

• Closure: For any two elements in the group, the result of their operation must also be in the group.
• Associativity: For any three elements in the group, the order of the operation does not matter. Mathematically, \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) for any elements \(a, b, c\) in the group.
• Identity Element: There must be an element in the group that does not change other elements when operated with them. Usually denoted as \(e\), it satisfies \(a \cdot e = a\) for any element \(a\).
• Inverse Element: For each element in the group, there must exist another element, known as its inverse, such that when both are operated, they yield the identity element, \(a \cdot a^{-1} = e\).

Understanding these foundations help unravel complex structures like normal subgroups and cosets.

Cosets

Cosets arise when investigating subgroups within a group. If you have a subgroup \(H\) of a group \(G\) and an element \(a\) from \(G\), you can form the left coset \(Ha\) and the right coset \(aH\).

• Left Coset \(Ha\): This is formed by taking each element \(h\) in the subgroup \(H\) and using the operation of the group to combine it with \(a\). So, \(Ha = \{ha \, | \, h \in H\}\).
• Right Coset \(aH\): Similarly, \(aH = \{ah \, | \, h \in H\}\).

In the context of normal subgroups, these cosets have a significant role. Specifically, when a subgroup \(H\) is normal, the left and right cosets are equal for every element in \(G\), i.e., \(Ha = aH\). This equality plays a crucial part in demonstrating the normality of the union of such cosets.

Closure Property

The closure property is crucial in proving that a set is a subgroup. To demonstrate a set \(S\), which is the union of cosets, as a subgroup, you must establish that it is closed under the group operation. Here are the steps:

• Understand \(S\): \(S\) is compiled from all elements of \(G\) for which the left and right cosets are identical—\(Ha = aH\).
• Select Elements: Choose any two elements \(x, y\) from \(S\). Because they belong to \(S\), each can be written as belonging to specific cosets \(Ha\) and \(Hb\).
• Combine Elements: Their product, \(xy\), needs to remain in a coset of the form \(Hab\), which satisfies the condition that this resulting coset is part of \(S\) through \((ab)H = H(ab)\).

This shows that the closure under multiplication holds for \(S\), a fundamental in proving \(S\) as a subgroup of \(G\).

Inverses in Groups

In group structures, every element must have an inverse. This is also necessary for proving a subgroup. For the set \(S\), which is the union of specific cosets, verifying the existence of inverses is essential.

• Existence of Inverses: For an element \(s\) in \(S\), we need its inverse \(s^{-1}\) to also be part of \(S\).
• Finding Inverses: If \(s\) is in a coset \(Ha\), then \(s^{-1}\) will be in \(Ha^{-1}\). This follows from the property \(Ha = aH\), leading to conjugate identities like \(H^{-1}a^{-1} = a^{-1}H\).
• Verification: Through this relationship, since \(Ha\) is equal to \(aH\), the inverse \(s^{-1}\) fits back into the set \(S\), confirming the subgroup property.

Making sure the inverses of all elements are in the set \(S\) completes the characterization of normal subgroups, ensuring all criteria for group properties are met.