Algebra

Group theory

The algebraic structure that captures symmetry: one operation, a handful of axioms, and a single idea running from the rotations of a cube to the impossibility of solving the quintic by radicals.

IntuitionWhat is symmetry?

Look at a square piece of paper. You can rotate it by 90°, 180°, 270°, or leave it alone, and it looks exactly the same. You can also flip it over four different ways. Each of these eight moves is a symmetry of the square: an action that leaves the shape looking unchanged. Doing one symmetry after another gives another symmetry — rotate 90° then flip, and the result is still one of the eight moves.

The rotations that map a cube to itself. Together with the identity, there are 24 of them — the same count as the ways to permute its four long diagonals.

This pattern — a set of moves, one way to combine them (do one after another), one move that does nothing, and every move has an undo — shows up everywhere in mathematics: symmetries of shapes, the ways to shuffle a deck, the non-zero rational numbers under multiplication, the invertible matrices under matrix product. Group theory is the study of exactly this pattern, stripped of what makes each example different.

SchoolClock arithmetic: a group hiding in plain sight

A clock has 12 hours: adding 5 hours to 10 o'clock gives 3 o'clock, because $10 + 5 = 15$ and $15$ "wraps around" to $3$ modulo 12. The set $\{0, 1, \dots, 11\}$ with this wrap-around addition is a group: adding 0 changes nothing, and every hour has an hour you can add to get back to 0 (5 and 7 undo each other). This is exactly the group $(\mathbb{Z}/12\mathbb{Z}, +)$ of congruences you may already know.

UndergraduateThe formal definition

Definition: Group

A group is a set $G$ together with an operation $*: G \times G \to G$ satisfying: (associativity) $(a*b)*c = a*(b*c)$ for all $a,b,c \in G$ ; (identity) there is $e \in G$ with $e*a = a*e = a$ for all $a$ ; (inverses) for every $a \in G$ 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.

(a * b) * c = a * (b * c), \qquad e * a = a * e = a, \qquad a * a^{-1} = a^{-1} * a = e

Examples: $(\mathbb{Z}, +)$ and $(\mathbb{Q}, +)$ are abelian groups with identity 0. $(\mathbb{Z}, \times)$ is not a group — most integers have no multiplicative inverse in $\mathbb{Z}$ — but $(\mathbb{R} \setminus \{0\}, \times)$ is a group. The invertible $n \times n$ matrices $\mathrm{GL}_n(\mathbb{R})$ form a group under matrix multiplication, and this one is not abelian once $n \ge 2$ : the order in which you compose two symmetries usually matters.

UndergraduateSubgroups, generators, and Cayley graphs

Definition: Subgroup

A subset $H \subseteq G$ is a subgroup if it contains $e$ and is closed under $*$ and under taking inverses — so $H$ is itself a group with the operation inherited from $G$ . A generating set is a subset $S \subseteq G$ such that every element of $G$ is a product of elements of $S$ and their inverses; a group generated by a single element is cyclic.

A Cayley graph turns a generating set into a picture: draw one vertex per element of $G$ , and join $g$ to $g*s$ for every generator $s$ . The 8 vertices below can be read as the group $(\mathbb{Z}/2\mathbb{Z})^3$ — triples of 0s and 1s added coordinate-wise mod 2 — generated by the three "flip one coordinate" moves. Every group has such a picture; it turns abstract multiplication into a graph you can walk around.

The cube graph

Q_3

: eight vertices, each joined to the three neighbours that differ in one coordinate. Read as a Cayley graph, it is

(\mathbb{Z}/2\mathbb{Z})^3

with generators

(1,0,0)

(0,1,0)

(0,0,1)

UndergraduateThe size of a subgroup

Theorem: Lagrange's theorem

If $G$ is a finite group and $H \le G$ is a subgroup, then $|H|$ divides $|G|$ .

Why is it true?

The left cosets $gH = \{gh : h \in H\}$ partition $G$ into blocks that all have exactly $|H|$ elements (the map $h \mapsto gh$ is a bijection $H \to gH$ ), so $|G|$ is $|H|$ times the number of cosets.

Proof

**Step 1 — Partition $G$ into left cosets via an equivalence relation.** Define $a \sim b \iff a^{-1}b \in H$ on $G$ . Because $H$ is a subgroup, it contains the identity ( $a^{-1}a = e \in H$ , reflexivity), is closed under inverses ( $(a^{-1}b)^{-1} = b^{-1}a \in H$ , symmetry), and is closed under products ( $(a^{-1}b)(b^{-1}c) = a^{-1}c \in H$ , transitivity). The equivalence class of $a$ is precisely the left coset $aH = \{ah : h \in H\}$ , so the distinct left cosets partition $G$ into $[G : H]$ disjoint subsets.

**Step 2 — Show every coset has size $|H|$ and sum the sizes.** For each $a \in G$ , the left-multiplication map $\phi_a : H \to aH,\; h \mapsto ah$ is surjective by definition and injective by left cancellation ( $ah_1 = ah_2 \implies h_1 = h_2$ after multiplying by $a^{-1}$ ). Hence $|aH| = |H|$ for every coset. Summing over all $[G : H]$ disjoint cosets gives $|G| = [G : H]\,|H|$ , proving that $|H|$ divides $|G|$ .

gH = \{g * h : h \in H\}, \qquad |G| = [G : H]\,|H|

Lagrange's theorem immediately explains why the rotation group of a cube (order 24, seen above) has subgroups only of order 1, 2, 3, 4, 6, 8, 12, 24 — the divisors of 24 — and never, say, order 5. It is a strong restriction: knowing $|G|$ alone already limits what subgroups can exist.

AdvancedSymmetric groups and why some equations have no formula

The set of all permutations of $n$ objects, composed by "do one then the other", is the symmetric group $S_n$ , of order $n!$ . The rotation group of the cube is (isomorphic to) $S_4$ : permuting the four long diagonals. Every finite group is a subgroup of some $S_n$ (Cayley's theorem), so symmetric groups already contain, in disguise, every possible finite symmetry.

Theorem: Cayley's theorem

Every group $G$ is isomorphic to a subgroup of the symmetric group $\mathrm{Sym}(G)$ of permutations of $G$ . In particular, every finite group of order $n = |G|$ embeds as a subgroup of $S_n$ .

Why is it true?

Abstract group axioms might look more general than concrete permutations, but Cayley's theorem proves they are not: left-multiplying $G$ by an element $g$ permutes the elements of $G$ faithfully, turning any abstract group into a concrete permutation group.

Proof

Step 1 — Associate a permutation to each group element. For each $g \in G$ , define the left-translation map $\lambda_g : G \to G,\; x \mapsto g * x$ . Because $\lambda_{g^{-1}}$ is its two-sided inverse, $\lambda_g$ is a bijection on $G$ , so $\lambda_g \in \mathrm{Sym}(G)$ .

**Step 2 — Verify that $\Lambda : G \to \mathrm{Sym}(G),\; g \mapsto \lambda_g$ is an injective homomorphism.** By associativity, $\lambda_{g * h}(x) = (g * h) * x = g * (h * x) = (\lambda_g \circ \lambda_h)(x)$ for all $x \in G$ , so $\Lambda(g * h) = \Lambda(g) \circ \Lambda(h)$ . If $\Lambda(g) = \mathrm{id}$ , evaluating at the identity gives $\lambda_g(e) = g * e = g = e$ , proving $\ker(\Lambda) = \{e\}$ . Thus $G$ is isomorphic to its image $\Lambda(G) \le \mathrm{Sym}(G)$ .

Quadratic, cubic and quartic equations all have formulas for their roots using $+, -, \times, \div$ and radicals. Whether a similar formula exists for the general quintic ( $n = 5$ ) turns out to be a question about the symmetric group $S_5$ : Galois theory attaches to every polynomial a group of symmetries of its roots, and a formula by radicals exists exactly when that group can be built up from abelian pieces (it is solvable). $S_5$ is not solvable — its subgroup $A_5$ has no normal subgroup other than itself and the trivial one — which is why no such formula can exist. This is the Abel–Ruffini theorem, and the full correspondence between fields and groups is the subject of `ly-thuyet-galois`.

UndergraduateReal-World Applications and Worked Examples

Group theory underpins modern public-key cryptography (Diffie–Hellman and elliptic-curve cryptography rely on cyclic groups of prime order $p$ where discrete logarithms are hard), error-correcting codes, spectroscopy and crystallography (point groups such as $D_4$ and $S_4$ classify molecular vibrations and crystal lattices), and particle physics.

Example: Cryptography: Primitive Roots and Inverses in $(\mathbb{Z}/11\mathbb{Z})^\times$

In the multiplicative group $(\mathbb{Z}/11\mathbb{Z})^\times$ used in Diffie–Hellman key exchange, use Lagrange's theorem to prove that $g = 2$ is a cyclic generator of the group and find the multiplicative inverse of $3$ .

Solution

Step 1 — Restrict candidate orders using Lagrange's theorem. The group $(\mathbb{Z}/11\mathbb{Z})^\times$ consists of $\{1, 2, \dots, 10\}$ under multiplication modulo $11$ , so $|G| = 10$ . By Lagrange's theorem, the order of any element must divide $10$ , leaving only $1, 2, 5, 10$ .

**Step 2 — Check the proper divisors for $g = 2$ .** Computing powers modulo $11$ : $2^1 \equiv 2$ , $2^2 \equiv 4$ , and $2^5 = 32 \equiv 10 \equiv -1 \not\equiv 1 \pmod{11}$ . Since no proper divisor of $10$ yields $1$ and $2^{10} \equiv (-1)^2 = 1 \pmod{11}$ , the order of $g = 2$ is $10$ , so $g = 2$ generates all of $(\mathbb{Z}/11\mathbb{Z})^\times$ .

**Step 3 — Find the inverse of $3$ .** Testing multiples of $3$ modulo $11$ , we find $3 \times 4 = 12 \equiv 1 \pmod{11}$ , so $3^{-1} \equiv 4 \pmod{11}$ .

Example: Molecular & Geometric Symmetry: The Dihedral Group $D_4$ of a Square

Let $D_4 = \langle r, s \mid r^4 = e,\; s^2 = e,\; srs = r^{-1} \rangle$ be the dihedral group of order $8$ generated by a $90^\circ$ rotation $r$ and a reflection $s$ . Verify that $D_4$ is non-abelian by simplifying $s r^2 s$ and $r s r$ , and list the left cosets of the rotation subgroup $H = \langle r \rangle = \{e, r, r^2, r^3\}$ .

Solution

**Step 1 — Use the commutation relation $srs = r^{-1} = r^3$ .** Multiplying on the left by $s$ (using $s^2 = e$ ) gives $r s = s r^3$ , so $r s \neq s r = r^3 s$ , confirming $D_4$ is non-abelian. Consequently, $r s r = (s r^3) r = s r^4 = s$ , and $s r^2 s = (srs)(srs) = r^{-1} r^{-1} = r^{-2} = r^2$ .

**Step 2 — Partition $D_4$ into left cosets of $H = \langle r \rangle = \{e, r, r^2, r^3\}$ .** Since $|D_4| = 8$ and $|H| = 4$ , Lagrange's theorem gives index $[D_4 : H] = 8 / 4 = 2$ . The two disjoint left cosets are the rotations $H = \{e, r, r^2, r^3\}$ and the reflections $sH = \{s, sr, sr^2, sr^3\}$ .

Which of the following is not a group under the given operation?

How many rotations (including doing nothing) map a cube onto itself?

By Lagrange's theorem, which of these can not be the order of a subgroup of a group of order 24?

Why does the general quintic equation ( $n = 5$ ) have no formula for its roots using $+, -, \times, \div$ and radicals?

References

David S. Dummit, Richard M. Foote (2004). Abstract Algebra
Michael Artin (2011). Algebra