Abstract Algebra — 01 · Groups and Subgroups

A group is a set with one operation that can always be undone. That short definition covers the integers under addition, clock arithmetic, the shuffles of a deck of cards, the rotations and reflections of a square, invertible matrices, and the unit quaternions used to rotate objects in 3D graphics. This post gives the axioms and their first consequences, builds the standard zoo of examples (symmetric, dihedral, matrix, quaternion and Klein four groups), and then studies subgroups, with a complete description of the subgroups of a cyclic group. It relies on post 00 for $\gcd$, the division algorithm and $\mathbb{Z}/n\mathbb{Z}$; post 02 builds on it with cosets and Lagrange's theorem.

Series map: 00 Integers, Relations & Modular Arithmetic → 01 Groups & Subgroups (this post) → 02 Cosets, Lagrange & Quotient Groups → 03 Homomorphisms, Group Actions & Sylow → 04 Rings, Ideals & Fields → 05 Field Extensions & Galois Theory.

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Learning Outcome

• State the group axioms and decide whether a given set and operation form a group.
• Prove the basic consequences: unique identity and inverses, cancellation, $(ab)^{-1} = b^{-1}a^{-1}$.
• Compute in $S_n$ with cycle notation, using a stated composition convention, and find the order and sign of a permutation.
• Work with the dihedral group $D_n$ through its generators $r, s$ and the relation $srs = r^{-1}$.
• Recognise $GL_n$, $SL_n$, the quaternion group $Q_8$ and the Klein four group, and build and check a Cayley table.
• Compute orders of elements, including $\lvert k \rvert = n/\gcd(k, n)$ in $\mathbb{Z}/n\mathbb{Z}$.
• Apply the one-step and two-step subgroup tests.
• Prove that every subgroup of a cyclic group is cyclic, list all subgroups of $\mathbb{Z}/n\mathbb{Z}$, and draw the subgroup lattice of $\mathbb{Z}/12\mathbb{Z}$.
• Form direct products and decide when $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ is cyclic.

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Chapter 1 — The group axioms

Terminology

• A binary operation on a set $G$ is a function $G \times G \to G$, written $(a, b) \mapsto a * b$ or just $ab$. "The result lies in $G$" is called closure; it is part of the definition of a binary operation.
• The operation is associative if $(ab)c = a(bc)$ for all $a, b, c$.
• An identity is an element $e$ with $ea = ae = a$ for all $a$.
• An inverse of $a$ is an element $b$ with $ab = ba = e$.
• A group is a set with an associative binary operation that has an identity, in which every element has an inverse.
• A group is abelian (commutative) if $ab = ba$ for all $a, b$. Abelian groups are often written additively: $a + b$, identity $0$, inverse $-a$.
• The order of a group, $\lvert G \rvert$, is the number of its elements (possibly infinite).

Examples and non-examples

Set and operation	Group?	Reason
$\mathbb{Z}$, $+$	yes, abelian	identity 0, inverse $-a$
$\mathbb{Z}$, $-$	no	not associative: $(1-1)-1 \neq 1-(1-1)$
$\mathbb{N}$, $+$	no	$1$ has no inverse in $\mathbb{N}$
$\mathbb{Z}$, $\times$	no	$2$ has no inverse in $\mathbb{Z}$
$\mathbb{Q}^\times = \mathbb{Q} \setminus \{0\}$, $\times$	yes, abelian	inverse $1/a$
$\mathbb{Z}/n\mathbb{Z}$, $+$	yes, abelian, order $n$	post 00
$\mathbb{Z}/6\mathbb{Z} \setminus \{0\}$, $\times$	no	not closed: $2 \cdot 3 = 0$
$(\mathbb{Z}/n\mathbb{Z})^\times$, $\times$	yes, abelian, order $\varphi(n)$	units, post 00
$n \times n$ real matrices, $\times$	no	singular matrices have no inverse

First consequences

Proposition 1.1. In any group $G$:

1. the identity is unique;
2. each element has exactly one inverse, written $a^{-1}$;
3. $(a^{-1})^{-1} = a$ and $(ab)^{-1} = b^{-1}a^{-1}$;
4. cancellation: $ab = ac$ implies $b = c$, and $ba = ca$ implies $b = c$;
5. the equations $ax = b$ and $ya = b$ have the unique solutions $x = a^{-1}b$ and $y = ba^{-1}$.

Proof. (1) If $e$ and $e'$ are identities, $e = ee' = e'$. (2) If $b$ and $c$ are inverses of $a$, then $b = be = b(ac) = (ba)c = ec = c$. (3) $a$ satisfies the definition of an inverse of $a^{-1}$, so by (2) it is the inverse. For the product, $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = e$, and similarly in the other order. (4) Multiply $ab = ac$ on the left by $a^{-1}$ and use associativity. (5) $x = a^{-1}b$ works; if $x'$ also works, $ax = ax'$ and cancellation gives $x = x'$. $\blacksquare$

Common pitfall: $(ab)^{-1} = b^{-1}a^{-1}$, not $a^{-1}b^{-1}$ (the "socks and shoes" rule: to undo putting on socks then shoes, remove the shoes first). The two agree only when $a$ and $b$ commute.

Powers are defined as usual: $a^0 = e$, $a^{n} = a \cdots a$ ($n$ factors), $a^{-n} = (a^{-1})^n$. The laws $a^m a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$ hold in every group, but $(ab)^n = a^n b^n$ can fail when $ab \neq ba$. In additive notation $a^n$ becomes $na$.

Cayley tables

A Cayley table lists every product $xy$, with $x$ labelling the row and $y$ the column. Part 5 of Proposition 1.1 says each element appears exactly once in every row and every column (a Latin square). Here is $(\mathbb{Z}/8\mathbb{Z})^\times = \{1, 3, 5, 7\}$ under multiplication mod 8:

$\times$	1	3	5	7
1	1	3	5	7
3	3	1	7	5
5	5	7	1	3
7	7	5	3	1

Every element squares to 1. Compare $(\mathbb{Z}/10\mathbb{Z})^\times = \{1, 3, 7, 9\}$: there $3^2 = 9$, $3^3 = 27 \equiv 7$, $3^4 = 81 \equiv 1$, so the powers of 3 already give every element. Both groups have order 4, but they are not "the same group": one has an element of order 4 and the other does not.

Watch out: a Latin square is necessary but not sufficient. A table can have every row and column a permutation and still fail associativity. Checking associativity by hand means checking $n^3$ triples, which is why groups are usually built from known ones (permutations, matrices, residues) where associativity is inherited.

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Chapter 2 — The standard examples

Symmetric groups $S_n$

A permutation of $\{1, \dots, n\}$ is a bijection from the set to itself. The permutations form the symmetric group $S_n$ under composition, with $\lvert S_n \rvert = n!$. Composition of functions is associative, the identity map is the identity, and a bijection has an inverse (post 00, Proposition 2.1).

Convention used in this series. $\sigma\tau$ means $\sigma \circ \tau$: apply $\tau$ first, then $\sigma$ (right to left, like function composition). Dummit & Foote, Gallian and Artin use this convention; some books and some software (including SymPy's Permutation product) use the opposite one, so always check.

Cycle notation. The cycle $(a_1 \ a_2 \ \dots \ a_k)$ sends $a_1 \mapsto a_2 \mapsto \dots \mapsto a_k \mapsto a_1$ and fixes everything else. It has length $k$; a 2-cycle is a transposition. Every permutation is a product of disjoint cycles, uniquely up to order and up to cycling each cycle's entries. Disjoint cycles commute.

To convert two-line notation: the permutation $1 \mapsto 3, 2 \mapsto 5, 3 \mapsto 1, 4 \mapsto 4, 5 \mapsto 6, 6 \mapsto 2$ follows $1 \to 3 \to 1$ and $2 \to 5 \to 6 \to 2$, with 4 fixed, so it is $(1 \ 3)(2 \ 5 \ 6)$.

Composing. In $S_3$ let $\sigma = (1 \ 2 \ 3)$ and $\tau = (1 \ 2)$. For $\sigma\tau$, follow each point through $\tau$ first, then $\sigma$: $1 \to 2 \to 3$, $2 \to 1 \to 2$, $3 \to 3 \to 1$. So $\sigma\tau = (1 \ 3)$. In the other order, $\tau\sigma$: $1 \to 2 \to 1$, $2 \to 3 \to 3$, $3 \to 1 \to 2$, so $\tau\sigma = (2 \ 3)$. Since $\sigma\tau \neq \tau\sigma$, $S_3$ is not abelian, and neither is $S_n$ for any $n \geq 3$.

A longer example in $S_5$: $(1 \ 3 \ 5)(1 \ 2)(3 \ 4)$. Apply $(3 \ 4)$, then $(1 \ 2)$, then $(1 \ 3 \ 5)$: $1 \to 1 \to 2 \to 2$; $2 \to 2 \to 1 \to 3$; $3 \to 4 \to 4 \to 4$; $4 \to 3 \to 3 \to 5$; $5 \to 5 \to 5 \to 1$. The product is the 5-cycle $(1 \ 2 \ 3 \ 4 \ 5)$.

Order of a permutation. A $k$-cycle has order $k$. For a product of disjoint cycles, the order is the least common multiple of the cycle lengths, because the cycles commute and act on separate points. So $(1 \ 2 \ 3)(4 \ 5)$ has order $\operatorname{lcm}(3, 2) = 6$, $(1 \ 3)(2 \ 5 \ 6)$ has order 6, and $(1 \ 2 \ 3 \ 4)(5 \ 6)$ has order 4.

Common pitfall: the lcm rule needs disjoint cycles. $(1 \ 2)(1 \ 3)$ is a product of two elements of order 2, but it equals $(1 \ 3 \ 2)$, which has order 3. Rewrite in disjoint form first.

The sign of a permutation

Every cycle is a product of transpositions: $(a_1 \ a_2 \ \dots \ a_k) = (a_1 \ a_k)(a_1 \ a_{k-1}) \cdots (a_1 \ a_2)$, which uses $k - 1$ transpositions. Hence every permutation is a product of transpositions, in many different ways. The number of transpositions is not fixed, but its parity is.

Theorem 2.1. Define $\operatorname{sgn} \sigma = \prod_{1 \leq i < j \leq n} \dfrac{\sigma(j) - \sigma(i)}{j - i}$. Then $\operatorname{sgn} \sigma \in \{+1, -1\}$, $\operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$, and every transposition has sign $-1$. Consequently a permutation cannot be both a product of an even number and of an odd number of transpositions.

Proof. The numerator runs over the same unordered pairs $\{\sigma(i), \sigma(j)\}$ as the denominator, each once, possibly with a sign change, so the product is $\pm 1$; it is $(-1)^{N}$ where $N$ is the number of inversions (pairs $i < j$ with $\sigma(i) > \sigma(j)$). For multiplicativity write

$$\operatorname{sgn}(\sigma\tau) = \prod_{i<j} \frac{\sigma(\tau(j)) - \sigma(\tau(i))}{\tau(j) - \tau(i)} \cdot \prod_{i<j} \frac{\tau(j) - \tau(i)}{j - i}.$$

The second product is $\operatorname{sgn} \tau$. In the first, each factor is unchanged when its two indices are swapped, and as $\{i, j\}$ runs over all pairs so does $\{\tau(i), \tau(j)\}$, so it equals $\operatorname{sgn} \sigma$. Finally, the transposition $(a \ b)$ with $a < b$ has exactly $2(b - a) - 1$ inversions: the pair $(a, b)$ and, for each $c$ strictly between them, the pairs $(a, c)$ and $(c, b)$. That number is odd. $\blacksquare$

So a $k$-cycle has sign $(-1)^{k-1}$: a 3-cycle is even, a transposition is odd. Examples: $\operatorname{sgn}((1 \ 2 \ 3 \ 4)(5 \ 6)) = (-1)^3 (-1)^1 = +1$; $\operatorname{sgn}((1 \ 3)(2 \ 5 \ 6)) = (-1)^1(-1)^2 = -1$. The even permutations form the alternating group $A_n$, of order $n!/2$ for $n \geq 2$.

Dihedral groups $D_n$

For $n \geq 3$, $D_n$ is the group of symmetries of a regular $n$-gon: rigid motions of the plane that map the polygon onto itself. Convention: in this series $D_n$ has order $2n$ (as in Gallian and Artin); Dummit & Foote call the same group $D_{2n}$.

Label the vertices $1, \dots, n$ counterclockwise. Let $r$ be rotation by $2\pi/n$ counterclockwise and $s$ the reflection in the line through vertex 1 and the centre. Then

$$D_n = \{e, r, r^2, \dots, r^{n-1}, \ s, rs, r^2 s, \dots, r^{n-1}s\}, \qquad r^n = s^2 = e, \quad srs = r^{-1}.$$

The $n$ elements $r^k$ are rotations and the $n$ elements $r^k s$ are reflections. The relation $srs = r^{-1}$ (equivalently $sr = r^{-1}s$) lets you move every $s$ to the right, so any word in $r$ and $s$ simplifies to one of the $2n$ normal forms. For example $s r^k = r^{-k} s$ and $(r^k s)^2 = r^k (s r^k s) = r^k r^{-k} = e$: every reflection has order 2.

Acting on the vertices, $D_3$ is all of $S_3$: $r = (1 \ 2 \ 3)$, $s = (2 \ 3)$, and then $r^2 = (1 \ 3 \ 2)$, $rs = (1 \ 2)$, $r^2 s = (1 \ 3)$. Its Cayley table (entry = row $\cdot$ column, column applied first):

$\cdot$	$e$	$r$	$r^2$	$s$	$rs$	$r^2s$
$e$	$e$	$r$	$r^2$	$s$	$rs$	$r^2s$
$r$	$r$	$r^2$	$e$	$rs$	$r^2s$	$s$
$r^2$	$r^2$	$e$	$r$	$r^2s$	$s$	$rs$
$s$	$s$	$r^2s$	$rs$	$e$	$r^2$	$r$
$rs$	$rs$	$s$	$r^2s$	$r$	$e$	$r^2$
$r^2s$	$r^2s$	$rs$	$s$	$r^2$	$r$	$e$

Row $s$, column $r$ gives $sr = r^2 s = r^{-1}s$, as the relation predicts. For the square, $D_4$ acting on vertices $1, 2, 3, 4$ has $r = (1 \ 2 \ 3 \ 4)$ and $s = (2 \ 4)$; its eight elements are four rotations of orders $1, 4, 2, 4$ and four reflections of order 2. Unlike $D_3$, $D_4$ is a proper subgroup of $S_4$ (order 8 versus 24).

Matrix groups

• $GL_n(\mathbb{R})$, the general linear group: invertible $n \times n$ real matrices under multiplication. Closure: $\det(AB) = \det A \det B \neq 0$.
• $SL_n(\mathbb{R})$, the special linear group: matrices of determinant 1.
• Over the field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$, $GL_2(\mathbb{F}_p)$ is finite. A matrix is invertible exactly when its columns are linearly independent: $p^2 - 1$ choices for a non-zero first column, then $p^2 - p$ for a second column that is not a multiple of the first. So $\lvert GL_2(\mathbb{F}_p) \rvert = (p^2 - 1)(p^2 - p)$, giving 6 for $p = 2$ and 48 for $p = 3$.

Matrix groups are the standard non-abelian examples: with $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$,

$$AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \qquad BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.$$

The quaternion group $Q_8$ and the Klein four group

$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ with $-1$ central (commuting with everything), $(-1)^2 = 1$, and Hamilton's rules

$$i^2 = j^2 = k^2 = ijk = -1, \qquad ij = k, \quad jk = i, \quad ki = j, \quad ji = -k, \quad kj = -i, \quad ik = -j.$$

Orders: $1$ has order 1, $-1$ has order 2, and the six elements $\pm i, \pm j, \pm k$ have order 4. $Q_8$ is non-abelian of order 8 but is not $D_4$: $D_4$ has five elements of order 2, $Q_8$ has only one.

The Klein four group $V = \{e, a, b, c\}$ has every non-identity element of order 2 and $ab = c$, $bc = a$, $ca = b$. The table of $(\mathbb{Z}/8\mathbb{Z})^\times$ above is $V$ with $a, b, c = 3, 5, 7$. Inside $S_4$ it appears as $\{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}$.

Groups from geometry: the circle and the roots of unity

• The circle group $S^1 = \{z \in \mathbb{C} : \lvert z \rvert = 1\}$ under multiplication: $\lvert zw \rvert = \lvert z \rvert \lvert w \rvert$ gives closure and $z^{-1} = \bar{z}$. Writing $z = e^{i\theta}$, multiplying means adding angles.
• The $n$-th roots of unity $\mu_n = \{e^{2\pi i k/n} : k = 0, \dots, n-1\}$ form a subgroup of $S^1$ of order $n$, generated by $\zeta = e^{2\pi i/n}$. They are the rotations of the regular $n$-gon, written as complex numbers.
• The orthogonal group $O(n)$ of real matrices with $A^{\mathsf{T}} A = I$ (distance-preserving linear maps) and its subgroup $SO(n)$ of rotations ($\det A = 1$). $SO(2)$ is the circle group again, with $e^{i\theta}$ matching the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.

These infinite groups are where group theory meets geometry and physics (Lie groups); this series stays mostly with finite groups.

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Chapter 3 — Orders of elements

Terminology

• The order $\lvert g \rvert$ of an element $g$ is the smallest $n \geq 1$ with $g^n = e$, or $\infty$ if no such $n$ exists. In additive notation: the smallest $n$ with $ng = 0$.
• The cyclic subgroup generated by $g$ is $\langle g \rangle = \{g^k : k \in \mathbb{Z}\}$.

In $\mathbb{Z}$ every non-zero element has infinite order. In $\mathbb{Z}/12\mathbb{Z}$, $4$ has order 3 ($4 + 4 + 4 = 12 \equiv 0$). In $\mathbb{R}^\times$, $-1$ has order 2 and $2$ has infinite order.

Lemma 3.1. Let $\lvert g \rvert = n < \infty$. Then $g^m = e$ if and only if $n \mid m$. Also $g^a = g^b$ if and only if $a \equiv b \pmod n$, so $\langle g \rangle = \{e, g, \dots, g^{n-1}\}$ has exactly $n$ elements.

Proof. Divide: $m = qn + r$ with $0 \leq r < n$. Then $g^m = (g^n)^q g^r = g^r$. By minimality of $n$, $g^r = e$ forces $r = 0$. The second claim applies the first to $g^{a-b}$. $\blacksquare$

So $\lvert g \rvert = \lvert \langle g \rangle \rvert$: the order of an element equals the order of the subgroup it generates. This is the link that lets Lagrange's theorem in post 02 say that element orders divide group orders.

Theorem 3.2. If $\lvert g \rvert = n$ and $k$ is an integer, then $\lvert g^k \rvert = \dfrac{n}{\gcd(k, n)}$. In particular, in $\mathbb{Z}/n\mathbb{Z}$ the class $k$ has order $n/\gcd(k, n)$.

Proof. Let $d = \gcd(k, n)$. By Lemma 3.1, $(g^k)^m = e$ iff $n \mid km$ iff $\frac{n}{d} \mid \frac{k}{d} m$. Since $\gcd(\frac{n}{d}, \frac{k}{d}) = 1$, Euclid's lemma turns this into $\frac{n}{d} \mid m$. The smallest positive such $m$ is $n/d$. $\blacksquare$

Examples in $\mathbb{Z}/20\mathbb{Z}$: $\lvert 8 \rvert = 20/\gcd(8, 20) = 20/4 = 5$, and $\langle 8 \rangle = \{0, 4, 8, 12, 16\} = \langle 4 \rangle$. The elements of order 20 are the $k$ with $\gcd(k, 20) = 1$: $1, 3, 7, 9, 11, 13, 17, 19$, which is $\varphi(20) = 8$ elements.

Proposition 3.3. In any group: (a) $\lvert g^{-1} \rvert = \lvert g \rvert$; (b) $\lvert ab \rvert = \lvert ba \rvert$; (c) if $ab = ba$ and $\gcd(\lvert a \rvert, \lvert b \rvert) = 1$ (both finite), then $\lvert ab \rvert = \lvert a \rvert \lvert b \rvert$.

Proof. (a) $(g^{-1})^m = (g^m)^{-1}$, which is $e$ exactly when $g^m = e$. (b) $ba = a^{-1}(ab)a$, and $(a^{-1} x a)^m = a^{-1} x^m a$ is $e$ exactly when $x^m = e$. (c) Let $m = \lvert a \rvert$, $n = \lvert b \rvert$. Since $a, b$ commute, $(ab)^{mn} = (a^m)^n (b^n)^m = e$. Conversely, if $(ab)^k = e$ then $a^k = b^{-k}$ lies in $\langle a \rangle \cap \langle b \rangle$. The order of this element divides both $m$ and $n$ (Lemma 3.1), so it is 1: $a^k = b^k = e$. Then $m \mid k$ and $n \mid k$, so $mn = \operatorname{lcm}(m, n)$ divides $k$. $\blacksquare$

Part (c) fails without commutativity: in $S_3$, $(1 \ 2)$ and $(1 \ 2 \ 3)$ have coprime orders 2 and 3, but their product $(1 \ 2)(1 \ 2 \ 3) = (2 \ 3)$ has order 2, not 6. It also fails without coprimality: in $\mathbb{Z}/4\mathbb{Z}$, $1$ and $3$ both have order 4, and $1 + 3 = 0$ has order 1.

Trick: $\langle k \rangle = \langle \gcd(k, n) \rangle$ in $\mathbb{Z}/n\mathbb{Z}$. Both have order $n/\gcd(k, n)$, and $\gcd(k, n)$ is a multiple of $k$ modulo $n$ by Bézout, so one contains the other and they have the same size.

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Chapter 4 — Subgroups

Terminology

• A subgroup $H \leq G$ is a subset that is a group under the same operation.
• $\{e\}$ is the trivial subgroup; $H \neq G$ is a proper subgroup.

The subgroup tests

Theorem 4.1 (two-step test). A non-empty subset $H \subseteq G$ is a subgroup if and only if it is closed under products ($a, b \in H \Rightarrow ab \in H$) and under inverses ($a \in H \Rightarrow a^{-1} \in H$).

Theorem 4.2 (one-step test). A non-empty subset $H \subseteq G$ is a subgroup if and only if $a, b \in H \Rightarrow ab^{-1} \in H$.

Theorem 4.3 (finite test). A non-empty finite subset $H \subseteq G$ closed under products is a subgroup.

Proofs. Two-step: associativity is inherited from $G$; pick $a \in H$, then $a^{-1} \in H$ and $e = aa^{-1} \in H$. One-step: take $b = a$ to get $e = aa^{-1} \in H$; then $a^{-1} = ea^{-1} \in H$; then $ab = a(b^{-1})^{-1} \in H$, so the two-step test applies. The "only if" directions are clear. Finite test: for $a \in H$ the powers $a, a^2, a^3, \dots$ lie in the finite set $H$, so two coincide, $a^i = a^j$ with $i < j$, and $a^{j-i} = e$. If $a = e$ its inverse is itself. Otherwise $j - i \geq 2$ and $a^{-1} = a^{j-i-1}$ is a positive power of $a$, hence in $H$. $\blacksquare$

Examples.

• $n\mathbb{Z} = \{nk : k \in \mathbb{Z}\} \leq \mathbb{Z}$: $na - nb = n(a - b)$ (one-step test, additive form $a - b$).
• $SL_n(\mathbb{R}) \leq GL_n(\mathbb{R})$: $\det(AB^{-1}) = \det A / \det B = 1$.
• $A_n \leq S_n$: the product of even permutations is even, and $\sigma^{-1}$ has the same sign as $\sigma$.
• The rotations $\{e, r, \dots, r^{n-1}\} = \langle r \rangle \leq D_n$.
• $\{z \in \mathbb{C} : z^n = 1\}$, the $n$-th roots of unity, is a subgroup of $\mathbb{C}^\times$ of order $n$.

Non-examples. The odd integers (no 0). $\{e, (1 \ 2), (1 \ 3)\} \subseteq S_3$ (not closed: $(1 \ 2)(1 \ 3) = (1 \ 3 \ 2)$). $\mathbb{N}$ in $\mathbb{Z}$ (no inverses).

Proposition 4.4. The intersection of any collection of subgroups is a subgroup. The union of two subgroups is a subgroup only if one contains the other.

Proof. Intersection: one-step test in each subgroup at once. Union: if $a \in H \setminus K$ and $b \in K \setminus H$, then $ab \notin H$ (else $b = a^{-1}(ab) \in H$) and $ab \notin K$ (else $a \in K$). $\blacksquare$

For example $2\mathbb{Z} \cup 3\mathbb{Z}$ contains $2$ and $3$ but not $5$. The smallest subgroup containing a set $S$ is the intersection of all subgroups containing $S$, written $\langle S \rangle$ and called the subgroup generated by $S$. For instance $D_n = \langle r, s \rangle$.

Worked example: subgroups of $D_4$ of order 4

With $r = (1 \ 2 \ 3 \ 4)$ and $s = (2 \ 4)$ acting on the vertices of a square, the eight elements of $D_4$ are

rotations	$e$	$r = (1 \ 2 \ 3 \ 4)$	$r^2 = (1 \ 3)(2 \ 4)$	$r^3 = (1 \ 4 \ 3 \ 2)$
reflections	$s = (2 \ 4)$	$rs = (1 \ 2)(3 \ 4)$	$r^2 s = (1 \ 3)$	$r^3 s = (1 \ 4)(2 \ 3)$

Is $K = \{e, r^2, s, r^2 s\}$ a subgroup? It is finite, so by Theorem 4.3 closure is enough. Using $s r^k = r^{-k} s$ and $r^4 = e$: $r^2 \cdot r^2 = e$, $s \cdot s = e$, $r^2 s \cdot r^2 s = e$ (reflections have order 2), $r^2 \cdot s = r^2 s$, $s \cdot r^2 = r^{-2} s = r^2 s$, $r^2 \cdot r^2 s = s$, $r^2 s \cdot r^2 = r^2 r^{-2} s = s$, $s \cdot r^2 s = r^{-2} = r^2$ and $r^2 s \cdot s = r^2$. Every product lands in $K$, so $K \leq D_4$. Every non-identity element of $K$ has order 2, so $K$ is a Klein four group, not cyclic.

$D_4$ has exactly three subgroups of order 4: the cyclic $\langle r \rangle = \{e, r, r^2, r^3\}$, and two Klein four groups, $K$ and $\{e, r^2, rs, r^3 s\}$. The last one, written as permutations, is $\{e, (1 \ 3)(2 \ 4), (1 \ 2)(3 \ 4), (1 \ 4)(2 \ 3)\}$, the copy of $V$ inside $S_4$ from Chapter 2.

Common pitfall: a subset is not a subgroup just because it contains $e$ and is closed under inverses. $\{e, s, rs\}$ has both properties but $s \cdot rs = r^{-1} s s = r^{-1} = r^3$ is missing, and in any case a subgroup of $D_4$ cannot have 3 elements (post 02: the order of a subgroup divides the order of the group).

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Chapter 5 — Cyclic groups

A group $G$ is cyclic if $G = \langle g \rangle$ for some $g$, called a generator. Cyclic groups are abelian ($g^a g^b = g^{a+b} = g^b g^a$). Examples: $\mathbb{Z} = \langle 1 \rangle = \langle -1 \rangle$; $\mathbb{Z}/n\mathbb{Z} = \langle 1 \rangle$; $(\mathbb{Z}/10\mathbb{Z})^\times = \langle 3 \rangle$; the rotation subgroup $\langle r \rangle$ of $D_n$. Non-examples: $V$ and $(\mathbb{Z}/8\mathbb{Z})^\times$ (no element of order 4), $\mathbb{Q}$ under addition, every non-abelian group.

Two groups are isomorphic, $G \cong H$, if there is a bijection $\phi : G \to H$ with $\phi(ab) = \phi(a)\phi(b)$; post 03 studies such maps properly. Lemma 3.1 shows that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ via $g^k \mapsto [k]$, and an infinite cyclic group is isomorphic to $\mathbb{Z}$. So up to isomorphism, $\mathbb{Z}$ and the $\mathbb{Z}/n\mathbb{Z}$ are the only cyclic groups.

Subgroups of cyclic groups

Theorem 5.1. Every subgroup of a cyclic group is cyclic.

Proof. Let $G = \langle g \rangle$ and $H \leq G$. If $H = \{e\}$, then $H = \langle e \rangle$. Otherwise $H$ contains some $g^k$ with $k \neq 0$, and also its inverse $g^{-k}$, so it contains a positive power of $g$. Let $m$ be the smallest positive integer with $g^m \in H$ (well-ordering). Clearly $\langle g^m \rangle \subseteq H$. Conversely take $g^t \in H$ and divide: $t = qm + r$ with $0 \leq r < m$. Then $g^r = g^t (g^m)^{-q} \in H$, and minimality of $m$ forces $r = 0$. So $g^t = (g^m)^q \in \langle g^m \rangle$, and $H = \langle g^m \rangle$. $\blacksquare$

Applied to $\mathbb{Z}$: every subgroup of $\mathbb{Z}$ is $n\mathbb{Z}$ for a unique $n \geq 0$.

Theorem 5.2 (fundamental theorem of finite cyclic groups). Let $G = \langle g \rangle$ have order $n$. For every positive divisor $d$ of $n$, $G$ has exactly one subgroup of order $d$, namely $\langle g^{n/d} \rangle$. It has no subgroups of other orders.

Proof. By Theorem 3.2, $\lvert g^{n/d} \rvert = n / \gcd(n/d, n) = n/(n/d) = d$, so a subgroup of order $d$ exists. Now let $H$ be any subgroup. By the proof of Theorem 5.1, $H = \langle g^m \rangle$ where $m$ is the least positive exponent in $H$ (if $H$ is trivial it is the unique subgroup of order 1). Since $g^n = e \in H$, the same division argument shows $m \mid n$. Then $\lvert H \rvert = \lvert g^m \rvert = n/\gcd(m, n) = n/m$. So the order of $H$ divides $n$, and $H$ is determined by its order: $\lvert H \rvert = d$ forces $m = n/d$. $\blacksquare$

Corollary 5.3. In a cyclic group of order $n$, $g^k$ is a generator if and only if $\gcd(k, n) = 1$; there are $\varphi(n)$ generators. The number of elements of order $d$ (for $d \mid n$) is $\varphi(d)$, so $\sum_{d \mid n} \varphi(d) = n$.

The subgroup lattice of $\mathbb{Z}/12\mathbb{Z}$

The divisors of 12 are $1, 2, 3, 4, 6, 12$, so $\mathbb{Z}/12\mathbb{Z}$ has exactly six subgroups:

order $d$	subgroup $\langle 12/d \rangle$	elements	elements of order $d$
1	$\langle 0 \rangle$	0	0
2	$\langle 6 \rangle$	0, 6	6
3	$\langle 4 \rangle$	0, 4, 8	4, 8
4	$\langle 3 \rangle$	0, 3, 6, 9	3, 9
6	$\langle 2 \rangle$	0, 2, 4, 6, 8, 10	2, 10
12	$\langle 1 \rangle$	all 12	1, 5, 7, 11

The last column has $1 + 1 + 2 + 2 + 2 + 4 = 12$ entries, one for each element, matching $\sum_{d \mid 12} \varphi(d) = 12$. Containment follows divisibility: the subgroup of order $d_1$ lies inside the subgroup of order $d_2$ exactly when $d_1 \mid d_2$. Drawing each subgroup as a node, with a line up to each subgroup that contains it with no subgroup in between, gives the subgroup lattice (Hasse diagram):

The lattice of $\mathbb{Z}/12\mathbb{Z}$ is the divisor lattice of 12 turned into groups. Non-cyclic groups have richer lattices: $D_4$, of order 8, has ten subgroups (one of order 1, five of order 2, three of order 4, one of order 8), and two different subgroups of order 4 can exist, which Theorem 5.2 forbids for cyclic groups.

Watch out: "for every divisor $d$ of $\lvert G \rvert$ there is a subgroup of order $d$" is a special property of cyclic groups (and, as post 03 shows, of some other families). Post 02 shows it fails for the alternating group $A_4$, of order 12, which has no subgroup of order 6.

Cayley's theorem

Permutation groups are not just one example among many: every group is one.

Theorem 5.4 (Cayley). Every group $G$ is isomorphic to a subgroup of the group $\operatorname{Sym}(G)$ of all permutations of the set $G$. In particular a group of order $n$ is isomorphic to a subgroup of $S_n$.

Proof. For $g \in G$ define $\lambda_g : G \to G$ by $\lambda_g(x) = gx$. It is a bijection, with inverse $\lambda_{g^{-1}}$. Associativity gives $\lambda_g \lambda_h (x) = g(hx) = (gh)x = \lambda_{gh}(x)$, so $g \mapsto \lambda_g$ respects the operations (with the right-to-left convention for composing). It is injective, because $\lambda_g = \lambda_h$ gives $g = \lambda_g(e) = \lambda_h(e) = h$. So $G$ is isomorphic to its image $\{\lambda_g\}$, which is a subgroup of $\operatorname{Sym}(G)$ (closed under products by the formula above and under inverses by $\lambda_g^{-1} = \lambda_{g^{-1}}$). $\blacksquare$

Example: in $(\mathbb{Z}/8\mathbb{Z})^\times = \{1, 3, 5, 7\}$, multiplication by 3 swaps $1 \leftrightarrow 3$ and $5 \leftrightarrow 7$, which is the row of 3 in the Cayley table read as a permutation; multiplication by 5 is $1 \leftrightarrow 5$, $3 \leftrightarrow 7$. Each row of a Cayley table is exactly the permutation $\lambda_g$, which is another way to see why rows never repeat an entry.

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Chapter 6 — Direct products and abelian groups

The (external) direct product of groups $G$ and $H$ is $G \times H$ with the componentwise operation $(g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)$. Its identity is $(e, e)$, inverses are $(g^{-1}, h^{-1})$, and $\lvert G \times H \rvert = \lvert G \rvert \lvert H \rvert$. It is abelian exactly when both factors are.

Proposition 6.1. If $g$ and $h$ have finite orders, then $\lvert (g, h) \rvert = \operatorname{lcm}(\lvert g \rvert, \lvert h \rvert)$.

Proof. $(g, h)^k = (g^k, h^k) = (e, e)$ iff $\lvert g \rvert \mid k$ and $\lvert h \rvert \mid k$ (Lemma 3.1), iff $\operatorname{lcm}(\lvert g \rvert, \lvert h \rvert) \mid k$. $\blacksquare$

Theorem 6.2. $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ is cyclic if and only if $\gcd(m, n) = 1$.

Proof. Every element has order $\operatorname{lcm}(\lvert a \rvert, \lvert b \rvert)$, which divides $\operatorname{lcm}(m, n) = mn/\gcd(m, n)$. If $\gcd(m, n) > 1$, no element reaches order $mn$, so the group is not cyclic. If $\gcd(m, n) = 1$, then $(1, 1)$ has order $\operatorname{lcm}(m, n) = mn$ and generates. $\blacksquare$

So $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z}$ (this is the additive half of the Chinese remainder theorem from post 00), while $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \cong V$ is not cyclic. In $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ (order 24) the largest element order is $\operatorname{lcm}(4, 6) = 12$; for example $(2, 3)$ has order $\operatorname{lcm}(2, 2) = 2$.

Direct products describe all finite abelian groups. The fundamental theorem of finite abelian groups (stated here; proved in the references) says that every finite abelian group is isomorphic to a direct product of cyclic groups of prime-power order, uniquely up to the order of the factors. For order 8 this gives exactly three abelian groups: $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $(\mathbb{Z}/2\mathbb{Z})^3$. Together with the non-abelian $D_4$ and $Q_8$ these are all five groups of order 8, up to isomorphism.

Trick: to tell groups apart, count elements of each order; an isomorphism preserves orders. $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $(\mathbb{Z}/2\mathbb{Z})^3$, $D_4$ and $Q_8$ have $1, 3, 7, 5, 1$ elements of order 2 respectively.

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Practice Exercises

1. Show that $G = \mathbb{R} \setminus \{-1\}$ with $a * b = a + b + ab$ is an abelian group. What are the identity and the inverse of $a$?

Solution

Note $1 + a * b = (1 + a)(1 + b)$. Closure: if $a, b \neq -1$ then $(1+a)(1+b) \neq 0$, so $a * b \neq -1$. Associativity and commutativity follow from those of multiplication, since both $(a * b) * c$ and $a * (b * c)$ equal $(1+a)(1+b)(1+c) - 1$. Identity: $0$. Inverse: $(1 + a)(1 + a') = 1$ gives $a' = \frac{1}{1+a} - 1 = \frac{-a}{1+a}$. In fact $a \mapsto 1 + a$ is an isomorphism to $\mathbb{R}^\times$.

2. In $S_6$ write $\sigma = (1 \ 2 \ 3)(4 \ 5)$ and $\tau = (1 \ 4)(2 \ 6)$. Compute $\sigma\tau$ in disjoint cycle form, its order and its sign.

Solution

Apply $\tau$ first: $1 \to 4 \to 5$, $5 \to 5 \to 4$, $4 \to 1 \to 2$, $2 \to 6 \to 6$, $6 \to 2 \to 3$, $3 \to 3 \to 1$. So $\sigma\tau = (1 \ 5 \ 4 \ 2 \ 6 \ 3)$, a 6-cycle: order 6, sign $(-1)^5 = -1$. Check with signs: $\operatorname{sgn}\sigma = (+1)(-1) = -1$, $\operatorname{sgn}\tau = (+1)$, product $-1$.

3. Find the order of every element of $(\mathbb{Z}/15\mathbb{Z})^\times$. Is it cyclic?

Solution

The group is $\{1, 2, 4, 7, 8, 11, 13, 14\}$, order $\varphi(15) = 8$. Orders: $1$ has order 1; $4, 11, 14$ have order 2 ($16, 121, 196$ are all $\equiv 1$); $2, 7, 8, 13$ have order 4 (e.g. $2^2 = 4$, $2^4 = 16 \equiv 1$). No element has order 8, so it is not cyclic. (It is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.)

4. In $D_n$ simplify $s r^3 s r^{2}$ and show that $r^k s$ has order 2 for every $k$.

Solution

$s r^3 s = r^{-3}$ (apply $srs = r^{-1}$ three times, or use $s r^k = r^{-k} s$ and $s^2 = e$). So $s r^3 s r^2 = r^{-3} r^2 = r^{-1}$. For the order: $(r^k s)(r^k s) = r^k (s r^k) s = r^k r^{-k} s s = e$, and $r^k s \neq e$.

5. List all subgroups of $\mathbb{Z}/20\mathbb{Z}$ with a generator for each, and find all generators of the subgroup of order 5.

Solution

One per divisor of 20: $\langle 0 \rangle$ (order 1), $\langle 10 \rangle$ (2), $\langle 5 \rangle$ (4), $\langle 4 \rangle$ (5), $\langle 2 \rangle$ (10), $\langle 1 \rangle$ (20). The subgroup of order 5 is $\{0, 4, 8, 12, 16\}$; all four non-zero elements generate it (each has order $20/\gcd(k, 20) = 20/4 = 5$), matching $\varphi(5) = 4$.

6. Show that $Q_8$ is not isomorphic to $D_4$, and that $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/24\mathbb{Z}$.

Solution

Isomorphisms preserve element orders. $Q_8$ has one element of order 2 ($-1$); $D_4$ has five ($r^2$ and the four reflections). $\mathbb{Z}/24\mathbb{Z}$ has an element of order 24; every element of $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ has order dividing $\operatorname{lcm}(4, 6) = 12$.

7. Prove that a group in which every element satisfies $g^2 = e$ is abelian.

Solution

Every element is its own inverse. For any $a, b$: $ab = (ab)^{-1} = b^{-1}a^{-1} = ba$.

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Cheat-sheet

Idea	Statement
Group	associative operation, identity, inverses
Inverse of product	$(ab)^{-1} = b^{-1}a^{-1}$
Cayley table	every row and column is a permutation
Composition in $S_n$	$\sigma\tau$ = apply $\tau$ first
Order of a permutation	lcm of disjoint cycle lengths
Sign	$k$-cycle has sign $(-1)^{k-1}$; $\lvert A_n \rvert = n!/2$
$D_n$ (order $2n$)	$r^n = s^2 = e$, $srs = r^{-1}$
$Q_8$	$i^2 = j^2 = k^2 = ijk = -1$
$GL_2(\mathbb{F}_p)$	order $(p^2-1)(p^2-p)$
Order and powers	$g^m = e$ iff $\lvert g \rvert \mid m$
Order of a power	$\lvert g^k \rvert = n/\gcd(k, n)$
One-step test	non-empty and $ab^{-1} \in H$
Cyclic subgroups	all cyclic; one per divisor $d$, namely $\langle g^{n/d} \rangle$
Generators of $\mathbb{Z}/n\mathbb{Z}$	$k$ with $\gcd(k, n) = 1$; $\varphi(n)$ of them
Direct product	$\lvert (g, h) \rvert = \operatorname{lcm}(\lvert g \rvert, \lvert h \rvert)$
Cyclic product	$\mathbb{Z}/m \times \mathbb{Z}/n$ cyclic iff $\gcd(m, n) = 1$

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What's next

Post 02 asks how a subgroup sits inside its group. Translating a subgroup $H$ by an element gives a coset $gH$; the cosets partition $G$ into pieces of equal size, which proves Lagrange's theorem: $\lvert H \rvert$ divides $\lvert G \rvert$. That single fact explains Theorem 5.2's "divisors only" and re-proves Fermat and Euler from post 00. Then comes the question of when the cosets themselves form a group, which leads to normal subgroups and quotient groups such as $\mathbb{Z}/n\mathbb{Z}$, which turns out to be $\mathbb{Z}$ read modulo its subgroup $n\mathbb{Z}$.

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References

• David S. Dummit and Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004 — Chapters 1–2 (groups, dihedral and symmetric groups, subgroups, cyclic groups, lattices).
• Joseph A. Gallian, Contemporary Abstract Algebra, 10th edition, CRC Press, 2021 — Chapters 1–5 and 8 (groups, finite groups and subgroups, cyclic groups, permutation groups, direct products).
• Michael Artin, Algebra, 2nd edition, Pearson, 2011 — Chapter 2 (groups).
• John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Addison-Wesley, 2003 — Sections 1–11.
• Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), judsonbooks.org — chapters on groups, cyclic groups and permutation groups.
• MIT OpenCourseWare, 18.701 Algebra I (Fall 2010) — lectures on groups, subgroups and permutations.
• Keith Conrad, Expository papers on group theory (University of Connecticut) — notes on dihedral groups, the sign of a permutation and the quaternion group.

Note: this post is in progress. Planned additions: the subgroup lattice of $D_4$ drawn out, a proof of the fundamental theorem of finite abelian groups (or a pointer to it once post 03 has homomorphisms), and matrix representations of $Q_8$.

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