Chapter 06 - Cyclic Groups

Cyclic groups are the most transparent groups in the subject. They are generated by a single element, their subgroup structure is governed entirely by divisor arithmetic, and they furnish the bridge between number theory and abstract algebra. Many later results --- normal subgroups, quotients, classification of finitely generated abelian groups --- are first learned in the cyclic case before being generalized.

§6.1 Cyclic Groups: Definition and Examples

Definition 6.1 (Cyclic group). A group $G$ is cyclic if there exists an element $a \in G$ such that every element of $G$ is a power of $a$. One writes $G = \langle a \rangle$. In additive notation, every element has the form $na$ for some $n \in \mathbb{Z}$. The element $a$ is called a generator of $G$.

Explicitly, if $a$ has finite order $n$, then

and $|G| = n$. If $a$ has infinite order, then all powers $a^k$ ($k \in \mathbb{Z}$) are distinct and $G$ is infinite.

Example 6.1: The integers $\mathbb{Z}$

The group $(\mathbb{Z}, +)$ is cyclic with generator $1$ (or $-1$). Every integer $n$ can be written as $n \cdot 1$. This is the prototype of an infinite cyclic group.

Example 6.2: $\mathbb{Z}_n$

The group $(\mathbb{Z}_n, +)$ of integers modulo $n$ is cyclic with generator $\bar{1}$. Every element $\bar{k}$ satisfies $\bar{k} = k \cdot \bar{1}$. This is the prototype of a finite cyclic group of order $n$.

Example 6.3: Roots of unity $\mu_n$

Let $\zeta = e^{2\pi i/n}$. The set

$$\mu_n = \{1, \zeta, \zeta^2, \dots, \zeta^{n-1}\}$$

forms a cyclic group of order $n$ under multiplication in $\mathbb{C}^{*}$. The generator $\zeta$ is a primitive $n$th root of unity. Geometrically, $\mu_n$ consists of $n$ equally spaced points on the unit circle, and multiplication corresponds to rotation by $2\pi/n$.

Example 6.4: Non-cyclic abelian group

The Klein four-group $V_4 = \mathbb{Z}_2 \times \mathbb{Z}_2 = \{(0,0),(1,0),(0,1),(1,1)\}$ is abelian but not cyclic: every non-identity element has order $2$, so no single element generates the whole group. This is the standard warning that “abelian” and “cyclic” are distinct notions.

§6.2 Every Cyclic Group is Abelian

Theorem 6.2. Every cyclic group is abelian.

Proof of Theorem 6.2

Let $G = \langle a \rangle$. Then every element of $G$ has the form $a^m$ for some $m \in \mathbb{Z}$. For any $a^m, a^n \in G$:

$$a^m \cdot a^n = a^{m+n} = a^{n+m} = a^n \cdot a^m.$$

The key step $a^{m+n} = a^{n+m}$ uses the commutativity of integer addition. Since every pair of elements commutes, $G$ is abelian. $\blacksquare$

The proof is short, but its content is important: once a group is controlled by powers of one element, the group operation reduces to arithmetic on exponents, and integer arithmetic is commutative. The converse is false --- $V_4$ is abelian but not cyclic.

§6.3 Subgroups of Cyclic Groups Are Cyclic

Theorem 6.3. Every subgroup of a cyclic group is cyclic.

Proof of Theorem 6.3 (Division algorithm argument)

Let $G = \langle a \rangle$ and let $H \le G$. If $H = \{e\}$, then $H = \langle e \rangle$ is trivially cyclic. Assume $H \neq \{e\}$.

Consider the set of positive exponents:

$$S = \{m \in \mathbb{N} : a^m \in H\}.$$

Since $H$ contains a non-identity element $a^k$ (with $k \neq 0$), and $H$ is a group so $a^{-k} \in H$ as well, the set $S$ is nonempty (it contains $|k|$). By the well-ordering principle, $S$ has a least element $d$.

Claim: $H = \langle a^d \rangle$.

The inclusion $\langle a^d \rangle \subseteq H$ is clear since $a^d \in H$ and $H$ is closed under the group operation.

For the reverse, let $h \in H$. Then $h = a^m$ for some integer $m$. Apply the division algorithm: write

$$m = qd + r, \qquad 0 \le r < d.$$

Then

$$a^r = a^{m - qd} = a^m \cdot (a^d)^{-q}.$$

Since $a^m \in H$ and $a^d \in H$, we have $a^r \in H$. But $0 \le r < d$ and $d$ is the smallest positive element of $S$. Therefore $r = 0$, so $m = qd$ and $h = (a^d)^q \in \langle a^d \rangle$.

Hence $H = \langle a^d \rangle$. $\blacksquare$

This proof is a paradigm for the “division algorithm” style of argument: to show a set is generated by its least positive element, divide an arbitrary element by that least element and argue the remainder must vanish.

§6.4 Classification of Cyclic Groups

Theorem 6.4 (Classification). Let $G$ be a cyclic group.

• If $G$ is infinite, then $G \cong \mathbb{Z}$.
• If $G$ is finite of order $n$, then $G \cong \mathbb{Z}_n$.

Proof of Theorem 6.4

Let $G = \langle a \rangle$ and define

$$\varphi : \mathbb{Z} \to G, \qquad k \mapsto a^k.$$

This is a homomorphism: $\varphi(k + l) = a^{k+l} = a^k \cdot a^l = \varphi(k)\varphi(l)$. It is surjective by the definition of a cyclic group.

Case 1: $G$ is infinite. If $a^k = a^l$ with $k \neq l$, then $a^{k-l} = e$ with $k - l \neq 0$, contradicting the assumption that $a$ has infinite order. So $\varphi$ is injective, hence an isomorphism $\mathbb{Z} \xrightarrow{\;\sim\;} G$.

Case 2: $G$ is finite of order $n$. The kernel of $\varphi$ is

$$\ker\varphi = \{k \in \mathbb{Z} : a^k = e\} = n\mathbb{Z},$$

since $a$ has order $n$. By the First Isomorphism Theorem (or by direct verification at this stage), $\varphi$ induces a bijective homomorphism

$$\bar{\varphi} : \mathbb{Z}/n\mathbb{Z} \xrightarrow{\;\sim\;} G, \qquad \bar{k} \mapsto a^k.$$

Therefore $G \cong \mathbb{Z}_n$. $\blacksquare$

This classification says that, up to isomorphism, there are exactly two kinds of cyclic group: $\mathbb{Z}$ and $\mathbb{Z}_n$. Every cyclic group is completely determined by the order of its generator.

§6.5 Order of Elements in $\mathbb{Z}_n$

Theorem 6.5 (Order formula). In $\mathbb{Z}_n$, the order of $\bar{k}$ is

Proof of Theorem 6.5

Let $d = \gcd(k, n)$. Write $k = dk_1$ and $n = dn_1$ where $\gcd(k_1, n_1) = 1$.

Step 1 (the order divides $n/d$). Compute:

$$\frac{n}{d} \cdot \bar{k} = \overline{\frac{n}{d} \cdot k} = \overline{n_1 \cdot dk_1} = \overline{k_1 n} = \bar{0} \pmod{n}.$$

So the order of $\bar{k}$ divides $n/d = n_1$.

Step 2 (the order is exactly $n/d$). Suppose $m \cdot \bar{k} = \bar{0}$, i.e., $n \mid mk$. Then $dn_1 \mid mdk_1$, which gives $n_1 \mid mk_1$. Since $\gcd(n_1, k_1) = 1$, Euclid’s lemma forces $n_1 \mid m$. Therefore every positive integer killing $\bar{k}$ is a multiple of $n_1 = n/d$.

Hence $\operatorname{ord}(\bar{k}) = n/d = n/\gcd(k,n)$. $\blacksquare$

Example 6.5a: Orders in $\mathbb{Z}_{12}$

Using $\operatorname{ord}(\bar{k}) = 12/\gcd(k,12)$:

$\bar{k}$	$\gcd(k,12)$	$\operatorname{ord}(\bar{k})$
$\bar{0}$	$12$	$1$
$\bar{1}$	$1$	$12$
$\bar{2}$	$2$	$6$
$\bar{3}$	$3$	$4$
$\bar{4}$	$4$	$3$
$\bar{5}$	$1$	$12$
$\bar{6}$	$6$	$2$
$\bar{7}$	$1$	$12$
$\bar{8}$	$4$	$3$
$\bar{9}$	$3$	$4$
$\bar{10}$	$2$	$6$
$\bar{11}$	$1$	$12$

§6.6 Generators of $\mathbb{Z}_n$ and Euler’s Totient Function

Theorem 6.6. The element $\bar{k}$ generates $\mathbb{Z}_n$ if and only if $\gcd(k, n) = 1$.

Proof of Theorem 6.6

By Theorem 6.5, $\operatorname{ord}(\bar{k}) = n/\gcd(k,n)$. The element $\bar{k}$ generates $\mathbb{Z}_n$ if and only if $\operatorname{ord}(\bar{k}) = n$, which occurs if and only if $\gcd(k,n) = 1$. $\blacksquare$

Definition 6.7 (Euler’s totient function). For $n \ge 1$, define

Equivalently, $\phi(n)$ is the number of generators of $\mathbb{Z}_n$.

Corollary 6.8. The cyclic group $\mathbb{Z}_n$ has exactly $\phi(n)$ generators.

Example 6.6a: Generators of $\mathbb{Z}_{20}$

The generators of $\mathbb{Z}_{20}$ are the elements $\bar{k}$ with $\gcd(k, 20) = 1$. Since $20 = 2^2 \cdot 5$, we need $k$ to be coprime to both $2$ and $5$:

$$\bar{1},\; \bar{3},\; \bar{7},\; \bar{9},\; \bar{11},\; \bar{13},\; \bar{17},\; \bar{19}.$$

There are $\phi(20) = 20 \cdot (1 - \tfrac{1}{2})(1 - \tfrac{1}{5}) = 20 \cdot \tfrac{1}{2} \cdot \tfrac{4}{5} = 8$ generators.

§6.7 Euler’s Totient Function: Formulas

Theorem 6.9. For a prime power $p^k$:

Proof of Theorem 6.9

Among the integers $1, 2, \dots, p^k$, the ones not coprime to $p^k$ are precisely the multiples of $p$:

$$p, 2p, 3p, \dots, p^{k-1} \cdot p.$$

There are $p^{k-1}$ such multiples. Therefore

$$\phi(p^k) = p^k - p^{k-1} = p^{k-1}(p-1). \quad \blacksquare$$

Theorem 6.10 (Multiplicativity). If $\gcd(m, n) = 1$, then

Proof of Theorem 6.10 (sketch via CRT)

By the Chinese Remainder Theorem, $\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{Z}_n$ when $\gcd(m,n) = 1$. An element $(\bar{a}, \bar{b})$ generates $\mathbb{Z}_m \times \mathbb{Z}_n$ as an additive cyclic group if and only if $\bar{a}$ generates $\mathbb{Z}_m$ and $\bar{b}$ generates $\mathbb{Z}_n$. The number of such pairs is $\phi(m) \cdot \phi(n)$, which must equal $\phi(mn)$. $\blacksquare$

Corollary 6.11 (General formula). If $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$, then

Example 6.7a: Computing $\phi(36)$

Since $36 = 2^2 \cdot 3^2$:

$$\phi(36) = 36 \cdot \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right) = 36 \cdot \frac{1}{2} \cdot \frac{2}{3} = 12.$$

Alternatively: $\phi(36) = \phi(4)\,\phi(9) = 2 \cdot 6 = 12$.

Example 6.7b: Computing $\phi(30)$

Since $30 = 2 \cdot 3 \cdot 5$:

$$\phi(30) = 30 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = 8.$$

§6.8 Subgroup Lattice of $\mathbb{Z}_n$

Theorem 6.12 (Subgroups of $\mathbb{Z}_n$). Let $n \ge 1$. For each divisor $d$ of $n$, there is exactly one subgroup of $\mathbb{Z}_n$ of order $d$, namely $\langle \overline{n/d} \rangle$. Conversely, every subgroup of $\mathbb{Z}_n$ has this form. In particular, the subgroups of $\mathbb{Z}_n$ are in bijection with the positive divisors of $n$.

Proof of Theorem 6.12

Existence. Let $d \mid n$ and set $m = n/d$. By Theorem 6.5,

$$\operatorname{ord}(\bar{m}) = \frac{n}{\gcd(m,n)} = \frac{n}{m} = d,$$

since $m \mid n$ implies $\gcd(m,n) = m$. So $\langle \bar{m} \rangle$ is a subgroup of order $d$.

Uniqueness. Let $H \le \mathbb{Z}_n$ with $|H| = d$. By Theorem 6.3, $H = \langle \bar{k} \rangle$ for some $k$. Then

$$d = \operatorname{ord}(\bar{k}) = \frac{n}{\gcd(k,n)},$$

so $\gcd(k,n) = n/d = m$. Write $k = ms$ with $\gcd(s, n/m) = \gcd(s, d) = 1$. Then $\bar{k} = s \cdot \bar{m}$, and since $\gcd(s,d) = 1$, the element $\bar{k}$ generates the same cyclic subgroup as $\bar{m}$. Hence $H = \langle \bar{m} \rangle = \langle \overline{n/d} \rangle$. $\blacksquare$

Worked Lattice: $\mathbb{Z}_{12}$

The divisors of $12$ are $1, 2, 3, 4, 6, 12$. The subgroups:

The containment relations are:

• $\langle\bar{6}\rangle \subset \langle\bar{3}\rangle \subset \langle\bar{1}\rangle = \mathbb{Z}_{12}$
• $\langle\bar{6}\rangle \subset \langle\bar{2}\rangle \subset \langle\bar{1}\rangle = \mathbb{Z}_{12}$
• $\langle\bar{4}\rangle \subset \langle\bar{2}\rangle$
• $\langle\bar{4}\rangle \subset \langle\bar{1}\rangle$
• $\langle\bar{3}\rangle \subset \langle\bar{1}\rangle$
• $\{\bar{0}\} \subset$ everything

Figure: subgroup lattice of $\mathbb{Z}_{12}$.

Figure: divisors, subgroups, and generators in $\mathbb{Z}_{12}$.

Read that figure row-by-row. For example, the divisor $4$ corresponds to the unique subgroup $\langle \bar{3}\rangle$ of order $4$, and that subgroup contributes exactly $\varphi(4)=2$ generators of order $4$, namely $\bar{3}$ and $\bar{9}$. This is the concrete mechanism behind both Theorem 6.12 and the identity $\sum_{d \mid n}\phi(d)=n$.

Worked Lattice: $\mathbb{Z}_{30}$

The divisors of $30 = 2 \cdot 3 \cdot 5$ are $1, 2, 3, 5, 6, 10, 15, 30$. The subgroups:

Figure: subgroup lattice of $\mathbb{Z}_{30}$.

Read it by divisor arithmetic: $\langle \overline{30/d_1}\rangle \subseteq \langle \overline{30/d_2}\rangle$ exactly when $d_1 \mid d_2$. For example, the order-$15$ subgroup $\langle \bar{2}\rangle$ contains the order-$5$ and order-$3$ subgroups, but not the order-$2$ subgroup.

The containment rule is: $\langle \overline{n/d_1} \rangle \subseteq \langle \overline{n/d_2} \rangle$ if and only if $d_1 \mid d_2$.

Worked Lattice: $\mathbb{Z}_{36}$

The divisors of $36 = 2^2 \cdot 3^2$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$. The subgroups:

The containment lattice is the divisibility lattice of $36$:

Here orders 18 and 12 are the maximal proper subgroups; their intersection is $\langle \bar{6} \rangle$ of order 6.

§6.9 The Identity $\sum_{d \mid n} \phi(d) = n$

Theorem 6.13. For every positive integer $n$,

Proof of Theorem 6.13 (partition of $\mathbb{Z}_n$ by generator order)

Consider the cyclic group $\mathbb{Z}_n$. For each element $\bar{k} \in \mathbb{Z}_n$, the order $\operatorname{ord}(\bar{k})$ is some divisor $d$ of $n$. The element $\bar{k}$ generates the unique subgroup of order $d$, which is isomorphic to $\mathbb{Z}_d$.

Now partition $\mathbb{Z}_n$ according to which subgroup each element generates. For each divisor $d$ of $n$, the unique subgroup of order $d$ has exactly $\phi(d)$ generators (the elements of $\mathbb{Z}_n$ whose order is exactly $d$).

Every element of $\mathbb{Z}_n$ generates exactly one cyclic subgroup, and its order equals the order of the element. Since the elements of order $d$ are precisely the generators of the unique subgroup of order $d$, and there are $\phi(d)$ of them, we obtain

$$n = |\mathbb{Z}_n| = \sum_{d \mid n} (\text{number of elements of order } d) = \sum_{d \mid n} \phi(d). \quad \blacksquare$$

Example 6.9a: Verification for $n = 12$

Divisors of $12$: $1, 2, 3, 4, 6, 12$.

$$\phi(1) + \phi(2) + \phi(3) + \phi(4) + \phi(6) + \phi(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12. \;\checkmark$$

§6.10 Worked Examples

Example 6.10a: All generators of $\mathbb{Z}_{20}$

We need all $\bar{k}$ with $1 \le k \le 20$ and $\gcd(k, 20) = 1$. Since $20 = 2^2 \cdot 5$:

The generators are:

Count: $\phi(20) = 20(1 - \tfrac{1}{2})(1 - \tfrac{1}{5}) = 8$. $\checkmark$

Example 6.10b: All subgroups of $\mathbb{Z}_{18}$

Since $18 = 2 \cdot 3^2$, the divisors of $18$ are $1, 2, 3, 6, 9, 18$.

There are exactly $6$ subgroups --- one for each divisor.

Example 6.10c: Order of every element in $\mathbb{Z}_{12}$

Using $\operatorname{ord}(\bar{k}) = 12/\gcd(k,12)$, here is the complete table:

Observe: the number of elements of each order matches $\phi(d)$:

• Order $1$: $1$ element ($\phi(1) = 1$)
• Order $2$: $1$ element ($\phi(2) = 1$)
• Order $3$: $2$ elements ($\phi(3) = 2$)
• Order $4$: $2$ elements ($\phi(4) = 2$)
• Order $6$: $2$ elements ($\phi(6) = 2$)
• Order $12$: $4$ elements ($\phi(12) = 4$)

Sum: $1+1+2+2+2+4 = 12$. $\checkmark$

§6.11 Lang’s Perspective: $\mathbb{Z}$ as the Free Cyclic Group

Lang’s point is stronger than the slogan “every cyclic group is either $\mathbb{Z}$ or $\mathbb{Z}_n$.” He begins from a universal property.

Theorem 6.14 (Universal property of $\mathbb{Z}$). Let $G$ be a group and let $g \in G$. Then there exists a unique group homomorphism

such that

In multiplicative notation this homomorphism is

Proof of Theorem 6.14

Existence. Define

$$\varphi_g(n) = g^n$$

for all integers $n$. We must check the homomorphism law:

$$\varphi_g(m+n) = g^{m+n} = g^m g^n = \varphi_g(m)\varphi_g(n).$$

So $\varphi_g$ is a homomorphism, and clearly $\varphi_g(1) = g$.

Uniqueness. Let $\psi : \mathbb{Z} \to G$ be any homomorphism with $\psi(1)=g$. Then for every positive integer $n$,

$$\psi(n)=\psi(\underbrace{1+\cdots+1}_{n\text{ times}})=\psi(1)^n=g^n.$$

Also,

$$e=\psi(0)=\psi(n+(-n))=\psi(n)\psi(-n),$$

so $\psi(-n)=\psi(n)^{-1}=g^{-n}$. Hence $\psi(n)=g^n$ for every integer $n$. Therefore $\psi=\varphi_g$. $\blacksquare$

This is the precise sense in which $\mathbb{Z}$ is “free on one generator”: once you specify where the generator $1$ goes, the entire homomorphism is forced.

Figure: the universal property of $\mathbb{Z}$ as the free cyclic group.

To specify a homomorphism out of $\mathbb{Z}$, it is enough to specify the image of $1$; the rest of the map is forced.

Corollary 6.15. The image of $\varphi_g$ is the cyclic subgroup generated by $g$:

Proof of Corollary 6.15

By definition,

$$\operatorname{im}(\varphi_g)=\{\varphi_g(n):n\in\mathbb{Z}\}=\{g^n:n\in\mathbb{Z}\}=\langle g\rangle.$$

So the cyclic subgroup generated by $g$ is not merely “like” the image of a map from $\mathbb{Z}$; it literally is that image. $\blacksquare$

Corollary 6.16. The kernel of $\varphi_g$ is:

• $\{0\}$ if $g$ has infinite order.
• $n\mathbb{Z}$ if $g$ has finite order $n$.

Proof of Corollary 6.16

By definition,

$$\ker(\varphi_g)=\{m\in\mathbb{Z}:g^m=e\}.$$

If $g$ has infinite order, the only such integer is $0$.

If $g$ has finite order $n$, then $g^m=e$ if and only if $n \mid m$. So

$$\ker(\varphi_g)=n\mathbb{Z}.$$

$\blacksquare$

Now the classification of cyclic groups becomes almost inevitable:

Corollary 6.17 (Classification reinterpreted). Every cyclic group is a quotient of $\mathbb{Z}$ by one of its subgroups. More precisely:

• if $g$ has infinite order, then $\langle g\rangle \cong \mathbb{Z}$;
• if $g$ has finite order $n$, then

Proof of Corollary 6.17

The image of $\varphi_g$ is $\langle g\rangle$ by Corollary 6.15. If $\ker(\varphi_g)=\{0\}$, then $\varphi_g$ is injective and $\langle g\rangle \cong \mathbb{Z}$.

If $\ker(\varphi_g)=n\mathbb{Z}$, then by the First Isomorphism Theorem,

$$\mathbb{Z}/n\mathbb{Z}\cong \operatorname{im}(\varphi_g)=\langle g\rangle.$$

$\blacksquare$

This packages several earlier facts into one picture:

• Subgroups of $\mathbb{Z}$ are exactly the kernels that can occur, so they must be of the form $n\mathbb{Z}$.
• Quotients of $\mathbb{Z}$ are exactly the cyclic groups.
• Finite versus infinite cyclic is controlled entirely by whether the chosen generator has nontrivial kernel.

Two concrete checks of the universal property

1. Take $G=\mathbb{Z}_{12}$ and $g=\bar{5}$. The unique homomorphism

   $$\varphi_{\bar{5}}:\mathbb{Z}\to\mathbb{Z}_{12}, \qquad n\mapsto \overline{5n}$$

   has image $\langle\bar{5}\rangle=\mathbb{Z}_{12}$ because $\gcd(5,12)=1$. Its kernel is $12\mathbb{Z}$ because $\bar{5}$ has order $12$.

2. Take $G=\mathbb{Z}_{12}$ and $g=\bar{4}$. Then

   $$\varphi_{\bar{4}}:\mathbb{Z}\to\mathbb{Z}_{12}, \qquad n\mapsto \overline{4n}$$

   has image

   $$\{\bar{0},\bar{4},\bar{8}\}=\langle\bar{4}\rangle$$

   and kernel $3\mathbb{Z}$ because $\bar{4}$ has order $3$. Therefore

   $$\mathbb{Z}/3\mathbb{Z}\cong \langle\bar{4}\rangle.$$

These examples are worth lingering over because they show the generator, the image, and the kernel all at once. In Lang’s style, a cyclic group is best understood not as a bare set with a generator, but as the image of the unique map out of the universal cyclic object $\mathbb{Z}$.

Bridge to Chapters 13 and 14 — from $\mathbb{Z}$ to homomorphisms and quotients

The universal-property section is where Chapter 6 stops being only about generators and starts becoming about maps.

Start with an element $g \in G$. The universal property gives a unique homomorphism

That one map already contains three later chapters in embryo:

• the image is the cyclic subgroup $\langle g\rangle$;
• the kernel records the order of $g$;
• the quotient $\mathbb{Z}/\ker(\varphi_g)$ is isomorphic to $\langle g\rangle$.

So the real structural route is

This becomes completely concrete when $g=\bar{1}\in \mathbb{Z}_n$. The corresponding homomorphism is the remainder map

Its image is all of $\mathbb{Z}_n$, its kernel is $n\mathbb{Z}$, and the First Isomorphism Theorem from Chapter 13 - Homomorphisms will say

Then Chapter 14 - Factor Groups reframes the same fact as a quotient-group construction: the residue classes modulo $n$ are the cosets of the normal subgroup $n\mathbb{Z}$ in $\mathbb{Z}$.

That is why Chapter 6 is much more than a list of examples of cyclic groups. It is the first place where the whole kernel-image-quotient pattern is already visible in a familiar setting.