The simplest way to build a subgroup is to start from one element and take everything it forces you to include. If $a \in H$, then all of $a, a^{2}, a^{3}, \dots$, $e$ and $a^{-1}, a^{-2}, \dots$ must be in $H$ too.
Definition. The cyclic subgroup generated by $a$ is $$\langle a \rangle = \{a^{k} : k \in \mathbb{Z}\}$$ (in additive notation $\langle a \rangle = \{ka : k \in \mathbb{Z}\}$). A group $G$ is cyclic if $G = \langle a \rangle$ for some $a$, called a generator.
Why it works
Cyclic groups are always abelian, since $a^{j}a^{k} = a^{j+k} = a^{k}a^{j}$. Examples: $\mathbb{Z} = \langle 1 \rangle$; $\mathbb{Z}_{n} = \langle 1 \rangle$; the rotations of a regular $n$-gon, generated by the smallest rotation; the $n$th roots of unity in $\mathbb{C}$, generated by $e^{2\pi i/n}$.
Which elements generate $\mathbb{Z}_{n}$?
Theorem. $k$ generates $\mathbb{Z}_{n}$ exactly when $\gcd(k, n) = 1$. Hence $\mathbb{Z}_{n}$ has $\varphi(n)$ generators.
Proof. If $\langle k \rangle = \mathbb{Z}_{n}$, then $1 \in \langle k \rangle$, so $kx \equiv 1 \pmod{n}$ for some $x$, and then $\gcd(k, n) = 1$. Conversely, if $\gcd(k, n) = 1$, Bézout gives $kx + ny = 1$, so $1 = xk$ lies in $\langle k \rangle$, and then every $m = m \cdot 1$ does. $\square$
Worked example: Generators of $\mathbb{Z}_{12}$
- Generators: the $k$ coprime to $12$: $1, 5, 7, 11$.
- $\langle 8 \rangle$: $8, 16 \equiv 4, 12 \equiv 0$, then it repeats. So $\langle 8 \rangle = \{0, 4, 8\}$, of order $3 = |8|$.
Answer: $1, 5, 7, 11$; and $\langle 8 \rangle = \{0, 4, 8\}$.
Worked example: Is $U(10)$ cyclic? Is $U(8)$?
- $U(10) = \{1, 3, 7, 9\}$: powers of $3$ are $3, 9, 7, 1$, all four elements. So $U(10) = \langle 3 \rangle$ is cyclic.
- $U(8) = \{1, 3, 5, 7\}$: $3^{2} = 9 \equiv 1$, $5^{2} = 25 \equiv 1$, $7^{2} = 49 \equiv 1$. Every element has order at most $2$, so none generates a group of order $4$.
Answer: $U(10)$ is cyclic; $U(8)$ is not.
Common mistake
Key idea