Problem 2 - A Book of Abstract Algebra

Question

Let \(G\) be a group, and \(H\) a normal subgroup of \(G\). Prove the following: If \((G: H)=m\), the order of every element of \(G / H\) is a divisor of \(m\).

Step-by-Step Solution

Answer

The order of every element of \(G/H\) divides \(m\).

Step 1: Understanding the Problem

We have a group \(G\) with a normal subgroup \(H\). We know that the index of \(H\) in \(G\) is \(m\), i.e., \((G:H) = m\). We need to show that for any element \(gH\) in the quotient group \(G/H\), the order of \(gH\) divides \(m\).

Step 2: Recall Definitions

The index \((G:H) = m\) means there are \(m\) distinct left cosets of \(H\) in \(G\). An element \(gH\) in the quotient group \(G/H\) has the form \(\{gh \mid h \in H\}\). The order of \(gH\) is the smallest positive integer \(n\) such that \((gH)^n = H\) in \(G/H\).

Step 3: Using Lagrange's Theorem

Lagrange's Theorem tells us that the order of a subgroup divides the order of the group. Here, \(H\) is normal, so the quotient group \(G/H\) is a well-defined group with order \(m\). Using the fact that \(gH\) is an element of \(G/H\), its order \(n\) will divide the order of \(G/H\), which is \(m\).

Step 4: Concluding the Proof

Hence, for any element \(gH\) in \(G/H\), the order \(n\) satisfies \((gH)^n = H\), and \(n\) divides the group order \(m\). Therefore, the order of every element of \(G/H\) is indeed a divisor of \(m\).

Key Concepts

Normal Subgroup

A normal subgroup, often denoted as \( H \), is a special type of subgroup within a group \( G \). A subgroup \( H \) is deemed normal if it satisfies the condition that for every element \( g \) in \( G \), the conjugate of \( H \) by \( g \), denoted \( gHg^{-1} \), is still \( H \). This is crucial because it allows us to form the quotient group \( G/H \), which is essential for solving many group theory problems involving symmetry and structure.

• Not every subgroup is normal, but when \( H \) is normal, it means the subgroup's structure is preserved even after transformation by elements from the larger group \( G \).
• Normality of a subgroup is central in defining cosets and constructing quotient groups.

Lagrange's Theorem

Lagrange's Theorem is a fundamental principle in group theory. It states that the order (or number of elements) of any subgroup \( H \) of a finite group \( G \) divides the order of \( G \). This theorem is significant because it helps us determine possible sizes of subgroups within a finite group.
For example, if \( G \) has 12 elements, any subgroup \( H \) can have an order that is any divisor of 12, such as 1, 2, 3, 4, 6, or 12. Importantly:

• In our problem, since \( H \) is a normal subgroup, the quotient group \( G/H \) can be formed, and its order is \( m \), the index of \( H \) in \( G \).
• According to Lagrange's Theorem, the order of any element in \( G/H \) must divide \( m \), because the order of each subgroup structure relates back to the full group \( G \).

Cosets

Cosets are key structures in understanding group theory. A coset is formed when you take a subgroup \( H \) and "shift" it by an element \( g \) from the larger group \( G \). There are two types of cosets. Left cosets, denoted \( gH \), and right cosets, denoted \( Hg \).

• In our context, when \( H \) is a normal subgroup, left and right cosets are identical, validating the construction of a quotient group \( G/H \).
• The number of distinct cosets (either left or right) is given by the index \( (G:H) = m \), highlighting how many times the subgroup \( H \) fits into the group \( G \).

By examining cosets, one gain insights on how groups partition into equal-sized, non-overlapping pieces, useful in many group-breaking techniques and symmetry analysis.