Classifying Finite Abelian Groups

Finite abelian groups can be listed completely: each one is a direct product of cyclic groups, in an essentially unique way. This is one of the great classification results of algebra, and it turns "how many abelian groups of order 72 are there?" into a counting exercise.

Fundamental theorem of finite abelian groups. Every finite abelian group is isomorphic to a product $$\mathbb{Z}_{p_{1}^{a_{1}}} \times \mathbb{Z}_{p_{2}^{a_{2}}} \times \cdots \times \mathbb{Z}_{p_{k}^{a_{k}}}$$ of cyclic groups of prime-power order (the primes need not be different). The list of prime powers, the elementary divisors, is unique up to order.

Why it works

Why prime powers, and why partitions? Two facts from the last lesson do most of the work. First, $\mathbb{Z}_{mn} \cong \mathbb{Z}_{m} \times \mathbb{Z}_{n}$ for coprime $m, n$, so any cyclic factor splits into prime-power pieces, and pieces for different primes never interact. Second, for one prime $p$, products like $\mathbb{Z}_{p^{2}} \times \mathbb{Z}_{p}$ and $\mathbb{Z}_{p^{3}}$ are not isomorphic (count the elements of order $p$). So for a group of order $p^{a}$ the possible shapes match the ways to write $a = a_{1} + a_{2} + \cdots$ with $a_{1} \ge a_{2} \ge \cdots$: the partitions of $a$. The existence part of the theorem (that every abelian group splits this way) takes a longer argument; Judson proves it in chapter 13.

Counting. If $n = p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}$, choose a partition of each exponent independently: the number of abelian groups of order $n$ is $\mathrm{p}(e_{1}) \cdots \mathrm{p}(e_{r})$, where $\mathrm{p}(e)$ is the number of partitions of $e$ ($\mathrm{p}(1) = 1$, $\mathrm{p}(2) = 2$, $\mathrm{p}(3) = 3$, $\mathrm{p}(4) = 5$, $\mathrm{p}(5) = 7$).

Worked example: Order 8

List the abelian groups of order $8 = 2^{3}$.

Partitions of $3$: $3$, $2 + 1$, $1 + 1 + 1$.
Groups: $\mathbb{Z}_{8}$, $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$, $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$.
They differ: the numbers of elements of order 2 are $1$, $3$ and $7$.

Answer: Three groups.

Worked example: Order 72

How many abelian groups of order $72$ are there? List them.

$72 = 2^{3} \cdot 3^{2}$: $\mathrm{p}(3) \cdot \mathrm{p}(2) = 3 \cdot 2 = 6$.
2-part: $\mathbb{Z}_{8}$, $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2}^{3}$; 3-part: $\mathbb{Z}_{9}$ or $\mathbb{Z}_{3} \times \mathbb{Z}_{3}$.
For example $\mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{9}$ is one of the six.

Answer: Six.

Worked example: Finding the elementary divisors

Write $\mathbb{Z}_{12} \times \mathbb{Z}_{18}$ with prime-power factors.

$\mathbb{Z}_{12} \cong \mathbb{Z}_{4} \times \mathbb{Z}_{3}$ and $\mathbb{Z}_{18} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{9}$.
Collect: $\mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{3}$.

Answer: Elementary divisors $4, 2, 9, 3$.

Common mistake

Counting $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ and $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$ as two groups: swapping factors gives an isomorphic group, so only the multiset of prime powers matters. The reverse slip is treating $\mathbb{Z}_{6}$ and $\mathbb{Z}_{2} \times \mathbb{Z}_{3}$ as different: they are isomorphic, since $\gcd(2, 3) = 1$. Always break factors into prime powers before comparing.

Key idea

Every finite abelian group is a product of cyclic groups of prime-power order, uniquely. For order $p^{a}$ the groups match the partitions of $a$; for general $n$, multiply the partition counts of the exponents.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 13, The Structure of Groups (section 13.1, Finite Abelian Groups)

MIT OpenCourseWare 18.702 Algebra II (Spring 2011), lecture notes on finite abelian groups

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 7.5, Finite Abelian Groups

Classifying Finite Abelian Groups

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