Prerequisites for Group Theory

Group theory is one of the most important tools in physics. This post collects the background needed to start learning it, and ends with the first few definitions and examples of groups.

Throughout, permutations are composed right to left: \(\sigma\tau\) means “apply \(\tau\) first, then \(\sigma\)”.

Equivalence relations

Why? Equivalence classes are needed to understand cosets, and hence important results such as Lagrange’s theorem.

Consider the natural numbers \(\mathbb{N}\) and their remainders on division by \(3\). Every number has remainder \(0\), \(1\) or \(2\), so we can group the numbers into three classes according to their remainder. Notice that every number lies in one of the classes, and no number lies in more than one.

In general, a relation \(\sim\) on a set \(S\) is an equivalence relation if it is:

• Reflexive: \(a \sim a\) for all \(a\). Non-example: “is smaller than” is not reflexive, because \(a<a\) is false.
• Symmetric: \(a \sim b\) implies \(b \sim a\). Non-example: “is less than or equal to” is not symmetric: \(5 \le 7\) but not \(7 \le 5\).
• Transitive: \(a \sim b\) and \(b \sim c\) imply \(a \sim c\). Non-example: in rock–paper–scissors, rock beats scissors and scissors beats paper, but rock does not beat paper.

The relation above is called congruence modulo 3. Other equivalence relations include:

• “is equal to” on the real numbers;
• “has the same real part as” on the complex numbers;
• “is similar to” on the set of triangles.

Partitions

Let \(A\) be a set and \(A_1, A_2, \ldots, A_n\) subsets of \(A\). They form a partition of \(A\) if

• their union is \(A\): \(\displaystyle\bigcup_{i=1}^n A_i=A\);
• none of them is empty: \(A_i \neq \varnothing\) for all \(i\);
• distinct subsets are disjoint: \(A_i \cap A_j=\varnothing\) for all \(i \neq j\).

The equivalence classes of any equivalence relation form a partition, and conversely every partition defines an equivalence relation.

Functions

Why? Bijections are at the heart of group theory: for example, the elements of the symmetric groups are bijections.

I assume familiarity with injections, surjections and bijections.

Exercise 1

Is \(f: \mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=x^2\), injective? Surjective? Bijective?

Solution. Not injective: \(f(1) = f(-1) = 1\). Not surjective: no real \(x\) has \(f(x) = -1\). Hence not bijective.

Exercise 2

Does \(x=y^2\) define \(y\) as a function of \(x\)?

Composition of bijections

Claim. If \(f: V \rightarrow V\) and \(g: V \rightarrow V\) are bijections, then so is \(f \circ g\).

Why it matters: it is what makes products of permutations such as \((1\,2\,3)(1\,2)=(1\,3)\) meaningful.

Proof.

Injective. Suppose \(f(g(v_1)) = f(g(v_2))\). Since \(f\) is injective, \(g(v_1) = g(v_2)\), and since \(g\) is injective, \(v_1 = v_2\).

Surjective. Let \(v \in V\). Since \(f\) is surjective there is \(v'' \in V\) with \(f(v'') = v\), and since \(g\) is surjective there is \(v' \in V\) with \(g(v') = v''\). Then \(f(g(v')) = v\).

Being injective and surjective, \(f \circ g\) is a bijection. \(\blacksquare\)

Graph isomorphisms and symmetries

An isomorphism between two graphs is a bijection between their vertex sets that preserves edges: two vertices are joined by an edge if and only if their images are. If graphs \(G\) and \(H\) are isomorphic we write \(G \cong H\). An isomorphism from a graph to itself is called a symmetry of the graph.

Groups

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

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

Theorem 1: the symmetries of a graph form a group

Why it matters: the dihedral group is exactly the set of symmetries of a regular \(n\)-gon.

Proof.

• (G1) Let \(f\) and \(g\) be symmetries. Then \(f \circ g\) is a bijection by the claim above, and \(v_1, v_2\) are joined by an edge \(\iff\) \(g(v_1), g(v_2)\) are joined \(\iff\) \(f(g(v_1)), f(g(v_2))\) are joined. So \(f \circ g\) is a symmetry.
• (G2) Composition of maps is associative: \((f \circ g) \circ h=f \circ(g \circ h)\). (Check this by evaluating both sides at a vertex.)
• (G3) The identity map \(e\) is a symmetry, and \(e \circ f = f \circ e =f\).
• (G4) A symmetry \(f\) is a bijection, so it has an inverse map \(f^{-1}\), and \(f^{-1}\) also preserves edges (try proving this by contradiction). Then \(f \circ f^{-1}=f^{-1} \circ f=e\).

Hence the symmetries of a graph form a group under composition. \(\blacksquare\)

Homomorphisms

A map \(\phi : G \rightarrow H\) between groups is a homomorphism if

\[\phi(x \circ y) = \phi(x) * \phi(y) \quad \text{for all } x, y \in G,\]

where \(\circ\) is the operation of \(G\) and \(*\) the operation of \(H\).

On the left, we combine \(x\) and \(y\) in \(G\) first and then apply \(\phi\). On the right, we apply \(\phi\) to each first and then combine the results in \(H\). A homomorphism is a map for which the two routes agree.

Example. Let \(G = (\mathbb{R}, +)\) and \(H = (\mathbb{R}_{>0}, \times)\), and take \(\phi = \exp\). Then \(\phi(x+y) = e^{x+y} = e^x e^y = \phi(x)\phi(y)\), so \(\exp\) is a homomorphism.

A homomorphism that is also a bijection is an isomorphism of groups, and we write \(G \cong H\). (The exponential map above is an isomorphism.)

Dihedral groups

The symmetries of a regular \(n\)-gon form a group by Theorem 1, called the dihedral group \(D_n\).

Theorem 2

The dihedral group \(D_n\) has exactly \(2n\) elements.

Proof. \(D_n\) consists of:

• the identity \(e\);
• the \(n-1\) anticlockwise rotations through angles \(\frac{2 \pi k}{n}\), \(k=1, \ldots, n-1\);
• \(n\) reflections (checking this is left as an exercise).

Since \(1+(n-1)+n=2n\), \(\lvert D_n\rvert=2 n\). \(\blacksquare\)

Exercise

Let \(g\) be the rotation through \(2\pi/n\) and \(h\) the reflection in the axis through a particular vertex. Show that

\[D_n=\left\{e, g, g^2, \ldots, g^{n-1}, h, g h, g^2 h, \ldots, g^{n-1} h\right\}.\]

Hint: what is the geometric meaning of \(g^{k} h\)?

Symmetric groups

The group \(S_n\) of all bijections of \(\{1, \ldots, n\}\) is the symmetric group, and \(\lvert S_n\rvert=n !\). It is also the symmetry group of the graph with \(n\) vertices and no edges. (Showing that \(S_n\) is a group is a good exercise.)

Matrix groups

Theorem 3

\(\mathrm{GL}(n, \mathbb{R})\), the set of invertible \(n\times n\) real matrices, is a group under matrix multiplication.

Proof.

• (G1) If \(A\) and \(B\) are invertible then \(\det A \neq 0\) and \(\det B \neq 0\), so \(\det(AB)=\det(A)\det(B) \neq 0\) and \(AB\) is invertible.
• (G2) Matrix multiplication is associative: \(A(BC)=(AB)C\) (exercise).
• (G3) The identity matrix \(I\) is the identity element.
• (G4) Every invertible \(A\) has an inverse \(A^{-1}\) with \(AA^{-1}=A^{-1}A=I\), and \(A^{-1}\) is itself invertible.

Hence \(\mathrm{GL}(n, \mathbb{R})\) is a group. \(\blacksquare\)

Subgroups

A subset \(H \subseteq G\) is a subgroup, written \(H \leq G\), if

1. \(H\) is not empty;
2. \(h, k \in H\) implies \(h * k \in H\);
3. \(h \in H\) implies \(h^{-1} \in H\).

A detour: \(\mathrm{SO}(3)\)

The rotation group \(\mathrm{SO}(3)\) is everywhere in physics.

In general,

\[\mathrm{SO}(n):=\left\{A \in \mathrm{GL}(n, \mathbb{R}) \;\middle|\; \det A=1 \text{ and } A^{T}=A^{-1}\right\},\]

the group of rotations of \(\mathbb{R}^n\): linear isometries that preserve orientation. It is the intersection \(\mathrm{SO}(n) = \mathrm{SL}(n, \mathbb{R}) \cap \mathrm{O}(n)\).

Exercise. Show that \(\mathrm{SO}(n)\) is a subgroup of \(\mathrm{GL}(n, \mathbb{R})\).

Abelian groups

A group \(G\) is abelian if \(g * h=h * g\) for all \(g, h \in G\).

Claim. \(D_3\) is not abelian.

Proof. Labelling the vertices of the triangle \(1, 2, 3\) identifies \(D_3\) with \(S_3\). Take

\[\sigma=\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}, \quad \tau=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}.\]

Composing right to left,

\[\sigma \tau=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, \quad \tau \sigma=\begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} .\]

Since \(\sigma \tau \neq \tau \sigma\), \(D_3\) is not abelian. \(\blacksquare\)

Normal subgroups

A subgroup \(N \leq G\) is normal, written \(N \trianglelefteq G\), if and only if \(g n g^{-1} \in N\) for all \(g \in G\) and \(n \in N\).

Theorem 4

Every subgroup of an abelian group is normal.

Proof. Let \(G\) be abelian and \(H \leq G\). For all \(g \in G\) and \(h \in H\), \(g h g^{-1}=g g^{-1} h= h \in H\). Hence \(H \trianglelefteq G\). \(\blacksquare\)

Cyclic groups

For \(g \in G\), define

\[\langle g\rangle:=\left\{g^k \mid k \in \mathbb{Z}\right\}=\left\{\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots\right\}.\]

A group (or subgroup) \(H\) is cyclic if \(H=\langle h\rangle\) for some \(h \in H\).

In the dihedral group \(D_n\), the subgroup of rotations

\[H=\left\{e, g, g^2, \ldots, g^{n-1}\right\} = \langle g\rangle\]

is cyclic. Verify that \(H\) is abelian; the same argument shows that every cyclic group is abelian.

Exercise. Explain why the converse fails: an abelian group need not be cyclic. (Hint: consider \(\mathbb{R}\) under addition.)

Cosets

For a subgroup \(H \leq G\) and \(g \in G\), the left coset of \(H\) by \(g\) is \(gH = \{gh : h \in H\}\).

Claim. \(g_1H=g_2H\) is possible even when \(g_1 \neq g_2\).

Proof. In \(S_3\), let \(H = \{e, (1\,2\,3), (1\,3\,2)\}\) be the subgroup of rotations. Then

\[\begin{aligned} (1\,2)H&=\{(1\,2),\ (1\,2)(1\,2\,3),\ (1\,2)(1\,3\,2)\} =\{(1\,2),(2\,3),(1\,3)\}, \\ (2\,3)H&=\{(2\,3),\ (2\,3)(1\,2\,3),\ (2\,3)(1\,3\,2)\}=\{(2\,3),(1\,3),(1\,2)\}, \end{aligned}\]

so \((1\,2)H = (2\,3)H\) although \((1\,2) \neq (2\,3)\). \(\blacksquare\)

In general, \(g_1H=g_2H\) if and only if \(g_1^{-1}g_2 \in H\). Here \((1\,2)^{-1}(2\,3) = (1\,2\,3) \in H\). So the map \(G \to \{\text{cosets of } H\}\), \(g \mapsto gH\), is not injective. (By contrast, left multiplication \(g \mapsto hg\) by a fixed element is always a bijection of \(G\).) The cosets of \(H\) partition \(G\) into equivalence classes of equal size, which is exactly the idea behind Lagrange’s theorem.

The alternating group

The even permutations in \(S_n\) form a subgroup, the alternating group \(A_n\). For \(n \ge 2\), exactly half of the permutations in \(S_n\) are even (composing with a fixed transposition swaps even and odd permutations), so \(\lvert A_n\rvert=\frac{n!}{2}\). For example, \(\lvert A_4\rvert = 12\).

Changes on 25 October 2026

What was wrong. Nothing in the results, but the convention for composing permutations was never stated, and one exercise solution only said “do it yourself”.

What changed. The composition convention is stated at the start, the solutions are written out, and the notation and wording are tidied. Previously titled “Pre-requisites for Group Theory”.