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?

  • $\begingroup$ (1) Yes, (2) Yes...but that may be heavily depending on the personal author's taste. $\endgroup$ – DonAntonio Dec 17 '17 at 23:18
  • $\begingroup$ 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". $\endgroup$ – eranreches Dec 17 '17 at 23:25
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    $\begingroup$ @DonAntonio (1) is also heavily dependent on the author's taste. ;) But of course, yes for OP's purposes. $\endgroup$ – Dustan Levenstein Dec 17 '17 at 23:26
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    $\begingroup$ 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 $\mathbb Z$. $\endgroup$ – Andres Mejia Dec 17 '17 at 23:27
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    $\begingroup$ See math.stackexchange.com/questions/792528/… $\endgroup$ – lhf Dec 17 '17 at 23:41

The smith normal form proof that Andres alludes to can be outlined as follows:

  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.
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    $\begingroup$ 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! $\endgroup$ – user491874 Dec 17 '17 at 23:53

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