Fundamental Theorem of Finite Abelian Groups

Theorem
Every finite abelian group is an internal group direct product of cyclic groups of prime-power order.

The number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.

Proof
Let $G$ be a finite abelian group.

By means of Abelian Group is Product of Prime-power Order Groups, we factor it uniquely into groups of prime-power order.

Then, Abelian Group of Prime-power Order is Product of Cyclic Groups applies to each of these factors.

Hence we conclude $G$ factors into prime-power order cyclic groups.

The factorisation of $G$ into prime-power order factors is already unique.

Therefore, a demonstration of the uniqueness of the secondary factorisation suffices:

Suppose $\left|{G}\right| = p^k$ with $p$ a prime.

Let $G$ have the following two factorisations:
 * $G = H_1 \times H_2 \times \cdots \times H_m = K_1 \times K_2 \times \cdots \times K_n$

where the $H_i$'s and $K_i$'s are non-trivial cyclic subgroups with
 * $\left|{H_1}\right| \ge \left|{H_2}\right| \ge \cdots \ge \left|{H_m}\right|$;
 * $\left|{K_1}\right| \ge \left|{K_2}\right| \ge \cdots \ge \left|{K_n}\right|$.

We proceed through induction on $k$.

Basis for the induction
For $k = 1$, the statement follows from Group of Prime Order Cyclic.

Induction Hypothesis
Now we assume the theorem is true for all abelian groups of order $p^l$, where $l < k$.

Induction Step
By Power of Elements is Subgroup, $G^p = \left\{{x^p: x \in G}\right\}$ is a proper subgroup of $G$.

It follows that:
 * $G^p = H_1^p \times \cdots \times H_{m'}^p=K_1^p \times \cdots \times K_{n'}^p$

where:
 * $m'$ is the largest integer $i$ such that $\left|{H_i}\right| > p$;
 * $n'$ is the largest integer $j$ such that $\left|{K_j}\right| > p$.

This $m'$ and $n'$ ensure that the direct products above do not have trivial factors.

Also, by Cauchy's Group Theorem, we have $\left|{G^p}\right| < \left|{G}\right|$.

This means that we can apply the induction hypothesis.

It follows that $m' = n'$ and $\left|{H_i^p}\right| = \left|{K_i^p}\right|$ for $i = 1, 2, \ldots, m'$.

We know that $\left|{H_i}\right| = p \left|{H_i^p}\right|$, and the same for $K_i$ and $K_i^p$.

It follows that $\left|{H_i}\right| = \left|{K_i}\right|$ for all $i=1, 2, \ldots, m'$.

Now by construction of $m'$, $i > m'$ implies $\left|{H_i}\right| = p = \left|{K_i}\right|$.

It follows that we have:
 * $\left|H_1\right| \left|H_2\right| \cdots \left|H_{m'}\right| p^{m-m'} = \left|G\right| = \left|K_1\right| \left|K_2\right| \cdots \left|K_{n'}\right| p^{n-n'}$

Therefore, $m - m' = n - n'$, and as $m' = n'$, it must be that $m = n$.

The result follows by induction.