Problem 37 - Algebra: Pure and Applied

Question

Let \(H\) be a subgroup of the group \(G\). Show that the map \(a \rightarrow a^{1}\) determines a one-to-one, onto map between the left cosets of \(H\) and the right cosets of \(H\).

Step-by-Step Solution

Answer

The map is bijective, proving a one-to-one, onto correspondence between left and right cosets.

Step 1: Understand the Problem

We need to prove that the map which assigns each left coset \( aH \) to its corresponding right coset \( Ha \) is both one-to-one (injective) and onto (surjective). A coset is a form of a subgroup \( H \) within a larger group \( G \), containing all elements \( gh \) for some element \( g \) in \( G \). Left coset is of the form \( aH = \{ ah \ | \ h \in H \} \), and right coset is of the form \( Ha = \{ ha \ | \ h \in H \} \).

Step 2: Define the Map

Define the map \( f: G/H \to H\backslash G \) by \( f(aH) = Ha \). We have to show that this map is well-defined, meaning that it does not depend on the representative of the coset.

Step 3: Show the Map is Well-defined

To show the map is well-defined, assume \( aH = bH \). This implies \( a^{-1}b \in H \). For the map \( f \), we want to show \( Ha = Hb \). Since \( a^{-1}b \in H \), we can write \( b = ah \) for some \( h \in H \). Then, \( Ha = H(ah) = Hb \), proving our map is well-defined.

Step 4: Prove Injectivity

To prove injectivity, assume \( f(aH) = f(bH) \), which means \( Ha = Hb \). This means there exists \( h_1 \in H \) such that \( ha = b \). Thus \( aH = bH \), proving that the map is injective.

Step 5: Prove Surjectivity

To prove surjectivity, for any right coset \( Ha \) in \( H \backslash G \), we need to find a left coset \( aH \) such that \( f(aH) = Ha \). Simply choose \( aH \), and it's clear that \( f(aH) = Ha \), showing surjectivity.

Key Concepts

Subgroups

In group theory, a **subgroup** is a smaller group contained within a larger group, obeying the group properties. If you have a group, say \( G \), and a set \( H \), then \( H \) is considered a subgroup of \( G \) if it satisfies two main conditions:

• Closure: For any two elements \( h_1 \) and \( h_2 \) in \( H \), their product \( h_1h_2 \) is also in \( H \).
• Inverse: For every element \( h \) in \( H \), its inverse \( h^{-1} \) is also in \( H \).

This concept is crucial as it maintains the structure and properties of the original group. It ensures that the subgroup can perform all the operations of the larger group and still remain within its own boundaries.

Cosets

A **coset** is formed by combining a subgroup with an element from the larger group. Imagine you have a group \( G \) and a subgroup \( H \). If you pick any element \( g \) from \( G \), a **left coset** of \( H \) in \( G \) is defined as \( gH = \{ gh \mid h \in H \} \). Conversely, a **right coset** is \( Hg = \{ hg \mid h \in H \} \).
Cosets help categorize group elements based on a subgroup. They partition a group into these smaller, non-overlapping subsets. This partitioning helps in analyzing the structure of groups and understanding how they operate through their subgroups.

Injective and Surjective Maps

In mathematics, particularly in group theory, functions or maps between sets can be classified as **injective** or **surjective**. Understanding these concepts is key to analyzing and proving properties like the bijection between left and right cosets:

• **Injective (one-to-one)**: A map is injective if different elements from the original set map to different elements in the target set. No two distinct elements in the original set share a target.
• **Surjective (onto)**: A map is surjective if every element of the target set has a pre-image in the original set. Meaning, the map covers the entire target set.

In our exercise, proving injectivity involves showing that if two left cosets map to the same right coset, then they are the same. Surjectivity involves showing every right coset has a left coset that maps to it.

Left and Right Cosets

The distinction between **left and right cosets** in a group is essential in understanding group structures and symmetries. Given a subgroup \( H \) of a group \( G \) and an element \( a \) from \( G \):

• A **left coset** is formed by multiplying \( a \) from the left with every element of \( H \), i.e., \( aH \).
• A **right coset** is formed by multiplying \( a \) from the right with every element of \( H \), i.e., \( Ha \).

These cosets are not always equal unless the subgroup is "normal." However, they are instrumental in this exercise, establishing a critical part of the proof by mapping each left coset to a right coset.

Well-defined Functions

In mathematics, a function is **well-defined** if it gives the same output for the same input, regardless of how the input is represented. In our scenario:

• Mapping each left coset \( aH \) to a right coset \( Ha \) needs consistency - it shouldn't matter which representative of the coset \( aH \) we choose to compute.
• This means, if for two elements \( a \) and \( b \) of the group, \( aH = bH \), then the resulting right cosets, \( Ha \) and \( Hb \), must also be equal.

Proving the function is well-defined is a critical part of ensuring that our mapping between cosets is valid and reliable, laying the foundation for more complex proofs in group theory.