Group Theory Basics: Groups, Subgroups and Cosets

Group theory is the mathematics of symmetry, and it is usually the first topic in an abstract algebra course. It can feel unfamiliar because you work with structures rather than numbers, but the core ideas are surprisingly concrete once you see examples. This guide covers groups, subgroups and cosets, and finishes with Lagrange's theorem.

What Is a Group?

A group \( (G, *) \) is a set \( G \) with a binary operation \( * \) satisfying four axioms:

1. Closure: for all \( a, b \in G \), \( a * b \in G \).
2. Associativity: \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in G \).
3. Identity: there is an element \( e \in G \) with \( e * a = a * e = a \) for all \( a \).
4. Inverses: for each \( a \in G \) there is \( a^{-1} \in G \) with \( a * a^{-1} = a^{-1} * a = e \).

If also \( a * b = b * a \) for all \( a, b \), the group is abelian (commutative).

Familiar Examples

• \( (\mathbb{Z}, +) \): identity 0, inverse of \( a \) is \( -a \).
• \( (\mathbb{R} \setminus \{0\}, \times) \): identity 1, inverse of \( a \) is \( \frac{1}{a} \).
• \( \mathbb{Z}_n = \{0, 1, \ldots, n-1\} \) under addition modulo \( n \).
• The six symmetries of an equilateral triangle, a non-abelian group called \( S_3 \).

Non-Examples

• \( (\mathbb{Z}, \times) \) is not a group: 2 has no integer inverse.
• \( (\mathbb{N}, +) \) is not a group: there are no negative inverses.

The order of a group, \( |G| \), is the number of its elements. The order of an element \( a \) is the smallest positive integer \( k \) with \( a^k = e \).

Worked Example 1: Verifying a Group

Show that \( U(8) = \{1, 3, 5, 7\} \) is a group under multiplication modulo 8.

Step 1: Closure. Compute every product modulo 8:

• \( 3 \times 3 = 9 \equiv 1 \)
• \( 5 \times 5 = 25 \equiv 1 \)
• \( 7 \times 7 = 49 \equiv 1 \)
• \( 3 \times 5 = 15 \equiv 7 \)
• \( 3 \times 7 = 21 \equiv 5 \)
• \( 5 \times 7 = 35 \equiv 3 \)

Products involving 1 are trivial. Every result lies in \( \{1, 3, 5, 7\} \), so the set is closed.

Step 2: Associativity. Multiplication of integers is associative, and reducing modulo 8 preserves this.

Step 3: Identity. \( 1 \times a = a \) for every \( a \), so \( e = 1 \).

Step 4: Inverses. From Step 1, each element squares to 1, so every element is its own inverse.

All four axioms hold, so \( U(8) \) is a group of order 4. Since multiplication is commutative, it is abelian. Interestingly, every non-identity element has order 2.

Subgroups

A subgroup \( H \) of \( G \) is a subset that is itself a group under the same operation. We write \( H \leq G \).

The One-Step Subgroup Test

A non-empty subset \( H \subseteq G \) is a subgroup if and only if

$$ a, b \in H \implies a * b^{-1} \in H $$

For a finite subset, it is even enough to check closure alone.

Every group has two trivial subgroups: \( \{e\} \) and \( G \) itself.

Cosets

Let \( H \leq G \) and \( a \in G \). The left coset of \( H \) containing \( a \) is

$$ aH = \{a * h : h \in H\} $$

In additive notation we write \( a + H \). Cosets have three key properties:

• Every coset has exactly \( |H| \) elements.
• Two cosets are either identical or completely disjoint.
• The cosets together cover all of \( G \).

So the cosets of \( H \) partition \( G \) into equal-sized pieces.

Worked Example 2: Subgroups and Cosets of \( \mathbb{Z}_6 \)

Let \( G = \mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\} \) under addition modulo 6.

Step 1: Find the subgroups. Generate subgroups from each element:

• \( \langle 0 \rangle = \{0\} \)
• \( \langle 1 \rangle = \langle 5 \rangle = \mathbb{Z}_6 \)
• \( \langle 2 \rangle = \{0, 2, 4\} \), since \( 2 + 2 = 4 \) and \( 4 + 2 = 6 \equiv 0 \)
• \( \langle 4 \rangle = \{0, 4, 2\} \), the same subgroup
• \( \langle 3 \rangle = \{0, 3\} \), since \( 3 + 3 = 6 \equiv 0 \)

So the subgroups have orders 1, 2, 3 and 6.

Step 2: Cosets of \( H = \{0, 3\} \).

$$ 0 + H = \{0, 3\} $$

$$ 1 + H = \{1, 4\} $$

$$ 2 + H = \{2, 5\} $$

Check: \( 3 + H = \{3, 0\} = 0 + H \), \( 4 + H = \{4, 1\} = 1 + H \), and \( 5 + H = \{5, 2\} = 2 + H \). So there are exactly 3 distinct cosets, each of size 2, and together they contain all 6 elements.

Step 3: Cosets of \( K = \{0, 2, 4\} \).

$$ 0 + K = \{0, 2, 4\}, \qquad 1 + K = \{1, 3, 5\} $$

Two cosets of size 3: the even and odd elements.

Lagrange's Theorem

The partition into equal cosets gives one of the most important results in algebra.

Lagrange's theorem: if \( G \) is a finite group and \( H \leq G \), then \( |H| \) divides \( |G| \), and

$$ [G : H] = \frac{|G|}{|H|} $$

where \( [G : H] \) is the number of distinct cosets, called the index.

In Example 2, \( \frac{6}{2} = 3 \) cosets of \( H \) and \( \frac{6}{3} = 2 \) cosets of \( K \), exactly as we found.

Useful Consequences

• The order of every element divides \( |G| \).
• A group of prime order \( p \) has no subgroups except \( \{e\} \) and \( G \), and it is cyclic.
• \( \mathbb{Z}_6 \) cannot have a subgroup of order 4 or 5.

Note that the converse fails: a group of order \( n \) need not have a subgroup of every order dividing \( n \).

Common Mistakes

• Forgetting to check every axiom, especially that the identity lies in the set.
• Assuming groups are abelian. In \( S_3 \), the order of operations matters.
• Thinking cosets are subgroups. Only the coset containing \( e \), namely \( H \) itself, is a subgroup. \( 1 + H = \{1, 4\} \) does not contain 0.
• Misusing Lagrange's converse. Divisibility does not guarantee a subgroup exists.

Key Takeaways

• A group needs closure, associativity, an identity and inverses.
• A subgroup is a subset that is a group under the same operation; use the one-step test.
• Cosets \( aH \) all have size \( |H| \) and partition \( G \).
• Lagrange's theorem: \( |H| \) divides \( |G| \), and the number of cosets is \( |G|/|H| \).

Working on an abstract algebra assignment or struggling with group theory proofs? Contact Amathify on WhatsApp or use the "Ask a Question" form, and our experts and PhD scholars will send step-by-step solutions.