How to find all subgroups of a cyclic group using the subgroup lattice?

**Subgroups of $\mathbb{Z}_{12}$ (additive, generated by 1 mod 12):** Divisors of 12: 1, 2, 3, 4, 6, 12. For each divisor $d$, the unique subgroup of order $d$ is generated by $12/d$: \begin{array}{c|c|c} d & \text{Generator} & \text{Subgroup} \\ \hline 1 & 12 & \langle 12 \rangle = \{0\} \\ 2 & 6 & \langle 6 \rangle = \{0, 6\} \\ 3 & 4 & \langle 4 \rangle = \{0, 4, 8\} \\ 4 & 3 & \langle 3 \rangle = \{0, 3, 6, 9\} \\ 6 & 2 & \langle 2 \rangle = \{0, 2, 4, 6, 8, 10\} \\ 12 & 1 & \langle 1 \rangle = \mathbb{Z}_{12} \end{array} **Subgroup Lattice Diagram:** \begin{matrix} & & \langle 1 \rangle \\ & \swarrow & \downarrow & \searrow \\ \langle 2 \rangle & & \langle 3 \rangle & & \langle 4 \rangle \\ \downarrow & \searrow & \swarrow & & \downarrow \\ \langle 6 \rangle & & \langle 0 \rangle \\ \searrow & \swarrow & & \\ & \langle 12 \rangle \text{ (not shown)} \end{matrix} Wait — the subgroup of order 6 ($\langle 2 \rangle$) contains the subgroup of order 3 ($\langle 4 \rangle$) only if 3 divides 6. Indeed $\langle 2 \rangle$ has elements 0, 2, 4, 6, 8, 10, and $\langle 4 \rangle = \{0, 4, 8\} \subset \langle 2 \rangle$. **Infinite Cyclic Group $\mathbb{Z}$:** Subgroups are $n\mathbb{Z} = \{\text{multiples of } n\}$ for $n \geq 0$. The lattice corresponds to divisibility: $m\mathbb{Z} \subseteq n\mathbb{Z}$ iff $n \mid m$. **Euler's Totient and Generators:** The number of generators of a cyclic group of order $n$ is $\varphi(n)$ — the number of integers $1 \leq k < n$ with $\gcd(k, n) = 1$. For $\mathbb{Z}_{12}$, $\varphi(12) = 4$: generators are $1, 5, 7, 11$. **Elements of order 6 in $\mathbb{Z}_6 \times \mathbb{Z}_6$:** An element $(a, b)$ has order 6 iff the lcm of the orders of $a$ and $b$ is 6. Count: elements of order 6 in $\mathbb{Z}_6$ are $a$ coprime to 6: $\varphi(6) = 2$ (elements 1 and 5). So $(a, b)$ has order 6 if: - $a$ has order 6 (2 choices) and $b$ has order 1, 2, 3, or 6: $1 + 1 + 2 + 2 = 6$ choices, plus the symmetric case, minus double-counting where both have order 6.