Finite abelian groups

In one sentence

Every finite abelian group is a direct sum of cyclic groups of prime power order, that decomposition is unique, and counting abelian groups of a given order is counting partitions of the prime exponents.

Why the exam asks it

Because it is one of the few places in algebra with a complete classification and an exact count, so questions can be precise: how many abelian groups of order $n$, are these two isomorphic, what is the exponent. And because $C_4\times C_2 \not\cong C_8$ is the small fact everything else rests on.

The idea

Fundamental theorem. Every finite abelian group is a direct sum of cyclic groups of prime power order, uniquely up to reordering.

Those prime powers are the elementary divisors. There is a second normal form:

$G \cong C_{d_1}\oplus\cdots\oplus C_{d_k}$ with $d_1 \mid d_2 \mid \cdots \mid d_k$,

the invariant factors, also unique. Each form determines the other, and questions are set in both, so converting between them is a required skill.

$C_4\oplus C_2 \oplus C_3 \oplus C_9$ has elementary divisors $4, 2, 3, 9$. To get invariant factors, put the prime powers in columns by prime, right-align, and multiply across rows:

$$\begin{array}{cc} 2 & 3\\ 4 & 9\end{array} \quad\Rightarrow\quad d_1 = 6,\; d_2 = 36,$$

so $G \cong C_6 \oplus C_{36}$, and indeed $6 \mid 36$.

Counting

The primes do not interact, so the count factors:

$$\#\{\text{abelian groups of order } p_1^{a_1}\cdots p_r^{a_r}\} = \prod_i p(a_i),$$

where $p(a)$ is the number of partitions of $a$.

For $72 = 2^3\cdot3^2$: $p(3)\cdot p(2) = 3 \times 2 = 6$. Only the exponents matter, so $5^3\cdot7^2$ also gives six.

When a product is cyclic

$$C_m \oplus C_n \cong C_{mn} \iff \gcd(m,n) = 1 .$$

That is the Chinese remainder theorem in group form, and it is the reason $C_2\oplus C_3 \cong C_6$ while $C_2\oplus C_2 \not\cong C_4$.

Consequently a finite abelian group is cyclic exactly when it has one invariant factor, equivalently when its elementary divisors use each prime at most once.

Exponent and orders

The exponent — the lcm of all element orders — is the largest invariant factor $d_k$, and an element of that order exists. So an abelian group is cyclic $\iff$ its exponent equals its order.

The number of elements of order dividing $p$ in an abelian $p$-group with $r$ elementary divisors for $p$ is $p^{r}$. So counting solutions of $x^p = e$ recovers $r$, which is how two groups of the same order are told apart: $C_8$ has one element of order $2$, $C_4\oplus C_2$ has three, and $C_2^3$ has seven.

Subgroups

Every subgroup and every quotient of a finite abelian group is again abelian, and — unusually — the converse of Lagrange holds: for every $d$ dividing $|G|$ there is a subgroup of order $d$. This is one of the few settings where that is safe, and it is worth knowing precisely because it fails in general.

Where intuition breaks

"$C_4\times C_2 \cong C_8$." Both have order $8$; only $C_8$ has an element of order $8$. Counting elements of order $2$ gives $3$ against $1$.

"$C_m\times C_n \cong C_{mn}$." Only when $\gcd(m,n) = 1$. This is the single most common slip in the topic.

"The number of abelian groups of order $n$ depends on $n$." Only on its exponents. $72$ and $200 = 2^3\cdot5^2$ both give $6$.

"There are $p(n)$ abelian groups of order $n$." Partitions of the exponents, multiplied — not of $n$. Order $12 = 2^2\cdot3$ gives $p(2)p(1) = 2$, not $p(12) = 77$.

"Elementary divisors and invariant factors are the same list." Different normal forms of the same group. $C_2\oplus C_4$ has elementary divisors $2,4$ and a single invariant-factor chain $2 \mid 4$; the lists coincide here and generally do not.

"An abelian group with an element of order $n$ has order $n$." It has exponent at least $n$. $C_2\oplus C_4$ has an element of order $4$ and order $8$.

"Every group of order $n$ with $n$ squarefree is cyclic." Every abelian one is. $S_3$ has order $6$, squarefree, and is not cyclic.

The exam's angle

1. Factor the order and partition each exponent. That is the complete answer to any counting question here.
2. To test isomorphism, count elements of each order — or convert both to invariant factors and compare, which is faster once the conversion is fluent.
3. Cyclic $\iff$ coprime factors $\iff$ one invariant factor $\iff$ exponent equals order. Four equivalent tests, and one of them is always cheap.
4. Watch whether "abelian" is stated. Several true statements here are false for groups in general, the converse of Lagrange most of all.
5. For a subgroup of given order, it exists — abelian groups are the safe case.

The night before

• Every finite abelian group is a direct sum of cyclic prime-power groups, uniquely.
• Two normal forms: elementary divisors (prime powers) and invariant factors ($d_1\mid\cdots\mid d_k$). Convert by right-aligning columns per prime and multiplying rows.
• Count $= \prod_i p(a_i)$, partitions of the exponents. Order $72 = 2^3 3^2$ gives $3\times2 = 6$.
• $C_m\oplus C_n \cong C_{mn} \iff \gcd(m,n)=1$.
• Cyclic $\iff$ one invariant factor $\iff$ exponent $=$ order.
• Exponent is the largest invariant factor, and is attained.
• Solutions of $x^p = e$ number $p^r$, with $r$ the count of $p$-elementary divisors. Order $8$: $1, 3, 7$ elements of order $2$ for $C_8$, $C_4\oplus C_2$, $C_2^3$.
• The converse of Lagrange holds for abelian groups. It does not in general.