# Does anyone know a simple proof for the fundamental theorem of finitely generated abelian group?

I am looking for a simple/short proof for the fundamental theorem of finitely generated abelian group.

Let $(G,\ +)$ be an abelian group.

Assume $G$ is finitely generated. Prove that there exist primes $p_{1}$, . . . $p_{r}$ and natural numbers $n, n_{i,j}$ such that

$G\simeq \mathbb{Z}^{n}\oplus(\mathbb{Z}/p_{1}^{n_{1,1}}\mathbb{Z}\oplus\cdots\oplus \mathbb{Z}/p_{1}^{n_{1,\mathrm{s}_{1}}}\mathbb{Z})\oplus\cdots\oplus(\mathbb{Z}/p_{r}^{n_{r,1}}\mathbb{Z}\oplus\cdots\oplus \mathbb{Z}/p_{r}^{n_{r,s_{r}}}\mathbb{Z})$ .

I also have difficulty with the notation:

1) Is this $\mathbb{Z}/p_{1}^{n_{1,1}}$ the same as $\mathbb{Z}_{p_{1}^{n_{1,1}}}$ ?

2) Is there an equivalent expression as the product instead of the direct sum?

• (1) Yes, (2) Yes...but that may be heavily depending on the personal author's taste. – DonAntonio Dec 17 '17 at 23:18
• It is a direct consequence of the classification of finitely generated modules over PID, though I'm not sure you'd consider this as a "proof". – eranreches Dec 17 '17 at 23:25
• @DonAntonio (1) is also heavily dependent on the author's taste. ;) But of course, yes for OP's purposes. – Dustan Levenstein Dec 17 '17 at 23:26
• If your linear algebra is strong, and you believe in the smith normal form, this essentially a proof for the structure theorem for modules over a PID which specializes to abelian groups for modules over $\mathbb Z$. – Andres Mejia Dec 17 '17 at 23:27
• – lhf Dec 17 '17 at 23:41

1. Every finitely generated abelian group $$G$$ is a quotient $$\mathbb Z^n/A$$ for $$A \subset \mathbb Z^n$$ an abelian subgroup, where $$n$$ is the size of a generating set for $$G$$.
2. Using the fact that every ideal of $$\mathbb Z$$ is principal, one can show by induction that $$A$$ itself is generated by at most $$n$$ elements.
3. So $$A$$ is given by $$\mathbb Z^n$$ modulo the image of $$M \in \operatorname{Mat}_{n \times m}(\mathbb Z)$$ with $$m \le n$$.
4. This gives us a description of an arbitrary abelian group as $$\mathbb Z^n$$ modulo the image of a matrix, and now we can choose to focus on what transformations on $$M$$ will preserve the isomorphism type of $$\mathbb Z^n/\operatorname{im}(M)$$. You can multiply $$M$$ on the right and left by any $$\mathbb Z$$-invertible matrix (i.e., $$\det=\pm 1$$), so this includes interchanging rows, adding a multiple of one row to another, and similarly for columns.
5. Put the smallest possible positive number in the upper left corner of $$M$$ using these row and column operations, and clear out the entries directly below and to its right. If the remaining $$(n-1) \times (m-1)$$ block is nonzero, then repeat the procedure inside that block. Note that minimality of the upper left entry implies that it divides all of the entries inside the $$(n-1) \times (m-1)$$ block.
6. The structure theorem that falls out of this procedure is $$G \simeq \mathbb Z/d_1 \mathbb Z \oplus \cdots \oplus \mathbb Z / d_k \mathbb Z$$ with $$d_1 \mid d_2$$, $$d_2 \mid d_3, \ldots, d_{k-1} \mid d_k$$. Now use the Chinese Remainder Theorem.
• I find this proof extremely instructive because it gives you a method of calculation as well. If you have relations between generators in $G$ you would write a system of Diophantine linear equations, and then solve it by reducing the matrix of the system - to Smith normal form! – user491874 Dec 17 '17 at 23:53