Subgroup Lattices

Subgroups are ordered by inclusion, a partial order, so we can draw them as a Hasse diagram: bigger subgroups higher, with a line when one sits directly inside another. This picture is called the subgroup lattice.

For $\mathbb{Z}_{n}$ the subgroups are $\langle d \rangle$ for the divisors $d$ of $n$, and $$\langle d \rangle \subseteq \langle d' \rangle \iff d' \mid d,$$ because $\langle d \rangle$ is the set of multiples of $d$. So the lattice of $\mathbb{Z}_{n}$ is the divisor diagram of $n$ turned upside down: $\langle 1 \rangle = \mathbb{Z}_{n}$ at the top, $\langle n \rangle = \{0\}$ at the bottom.

To draw it by hand, put $\{0\}$ at the bottom and climb one prime at a time: from $\langle d \rangle$ you can step up to $\langle d/p \rangle$ for each prime $p$ dividing $d$. Every edge then has index $p$, a prime, and the subgroups at each height have the same number of prime factors in their index.

The subgroup lattice of $\mathbb{Z}_{24}$. Tap $\langle 4 \rangle$: above it are $\langle 2 \rangle$ and $\langle 1 \rangle$ (generators that divide 4), and inside it are $\langle 8 \rangle$, $\langle 12 \rangle$ and $\{0\}$. The numbers on the edges are indices.

Why it works

For divisors $a$ and $b$ of $n$: $$\langle a \rangle \cap \langle b \rangle = \langle \operatorname{lcm}(a, b) \rangle, \qquad \text{the smallest subgroup containing both} = \langle \gcd(a, b) \rangle.$$ The intersection is the set of common multiples of $a$ and $b$, which are the multiples of the lcm. A subgroup containing $a$ and $b$ contains every $sa + tb$, and by Bézout those are the multiples of $\gcd(a, b)$. In the diagram these are the highest point below both and the lowest point above both.

Worked example: Reading the lattice of $\mathbb{Z}_{36}$

In $\mathbb{Z}_{36}$, find $\langle 6 \rangle \cap \langle 9 \rangle$ and the smallest subgroup containing both.

$\operatorname{lcm}(6, 9) = 18$, so $\langle 6 \rangle \cap \langle 9 \rangle = \langle 18 \rangle = \{0, 18\}$.
$\gcd(6, 9) = 3$, so the smallest subgroup containing both is $\langle 3 \rangle$, of order $12$.

Answer: $\langle 18 \rangle$ and $\langle 3 \rangle$.

Worked example: Directly above

Which subgroups of $\mathbb{Z}_{24}$ sit directly above $\langle 12 \rangle$?

$\langle d \rangle \supsetneq \langle 12 \rangle$ means $d$ is a proper divisor of $12$.
Nothing lies in between when $12/d$ is prime: $d = 6$ ($12/6 = 2$) and $d = 4$ ($12/4 = 3$).

Answer: $\langle 6 \rangle$ and $\langle 4 \rangle$.

Groups that are not cyclic have richer lattices. $\mathbb{Z}_{4}$ has a single subgroup of order $2$, but $U(8) = \{1, 3, 5, 7\}$, also of order $4$, has three: $\{1, 3\}$, $\{1, 5\}$ and $\{1, 7\}$. The symmetries of a square, $D_{4}$, have ten subgroups.

The ten subgroups of $D_{4}$, the eight symmetries of a square (you will meet it soon). Three subgroups have order 4, five have order 2.

Common mistake

Drawing $\langle 6 \rangle$ above $\langle 2 \rangle$ in $\mathbb{Z}_{12}$ because $6 \gt 2$. It is the other way around: $\langle 6 \rangle = \{0, 6\}$ is inside $\langle 2 \rangle = \{0, 2, 4, 6, 8, 10\}$. A larger generator means fewer multiples, so a smaller subgroup.

Key idea

The subgroup lattice shows inclusion. For $\mathbb{Z}_{n}$, $\langle d \rangle \subseteq \langle d' \rangle$ exactly when $d' \mid d$; intersections come from lcms and joins from gcds.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 4, Cyclic Groups (subgroup diagrams); chapter 19, Lattices and Boolean Algebras (section 19.1)

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 3.5, Cyclic Groups (subgroup diagrams)

Subgroup Lattices

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