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.

Por qué funciona

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 $e$ ($\mathrm{p}(1) = 1$, $\mathrm{p}(2) = 2$, $\mathrm{p}(3) = 3$, $\mathrm{p}(4) = 5$, $\mathrm{p}(5) = 7$).

Ejemplo resuelto: 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 $7$.

Respuesta: Three groups.

Ejemplo resuelto: 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.

Respuesta: Six.

Ejemplo resuelto: 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: $\mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{3}$.

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

Error común

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.

Idea clave

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.

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