While the literature has many variations, all the published proofs I know induct on the number of generators. Thus they start will an abelian group $A$ and build up a direct sum of cyclic groups within it. Part of the difficulty involves the occasional need to remove a direct summand and replace it with another in order to proceed.

My question: Does any published proof induct on the relations?

With no relations, one has a free abelian group, so a direct sum of infinite cyclic groups. Adding a relation induces a quotient, and the idea is to see how the direct sum decomposition behaves with respect to the quotient process.

To illustrate one important step, I'll explain how to argue that the torsion subgroup of a finitely generated abelian group has a free abelian complement. Assume this by induction and add a new relation. The relation has the form $f+t$, where $f$ belongs to the free abelian summand and $t$ to the torsion subgroup. Say $t$ has order $m$. Then $m(f+t)=mf$. If $f$ does not equal $0$, choose a basis for the free group that makes $mf$ a multiple of a basis element (standard stuff). Now modding out by $mf$ produces torsion and reduces the rank. Then modding out by $f$ takes a quotient of the torsion.

Analyzing the torsion part turns out only just a little more invovled.

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    $\begingroup$ @Rocket Man: I'm in the category of abelian groups, not the category of groups. The commutative law comes with the hardware. $\endgroup$ – David Feldman Mar 27 at 3:19
  • $\begingroup$ Is it clear a priori that a finitely generated abelian group is also finitely presented? $\endgroup$ – anomaly Mar 27 at 3:21
  • $\begingroup$ Yes. Certainly a free abelian group is finitely presented. Then if you follow my argument, each new relation either reduces the rank or preserves the rank and reduces the torsion. So this follows by double induction. $\endgroup$ – David Feldman Mar 27 at 3:23
  • $\begingroup$ @anomaly : $\mathbb{Z}$ is noetherian, so submodules of $\mathbb{Z}^r$ are finitely generated for finite $r$ $\endgroup$ – Max Mar 27 at 7:37

Sure, although maybe not explicitly. The proofs I know are geometric and essentially use the algorithm for Smith normal form. Basically, you surjectively map $f:\mathbb{Z}^n \twoheadrightarrow A$, then you map $g:\mathbb{Z}^m\to \mathbb{Z}^n$ so that $\text{im}(g) = \ker(f)$. So $m$ is the number of generators of the subgroup of relations. The cokernel of $g$ is isomorphic to $A$. If $\phi$ is an automorphism of $\mathbb{Z}^n$ and $\psi$ an automorphism of $\mathbb{Z}^m$, then $A$ is also isomorphic to the cokernel of $\phi \circ g \circ \psi^{-1}$, i.e., you can change bases without messing up the cokernel. Then you look at the matrix $M$ for $g$. The previous statement allows you to do certain matrix operations to clean up $M$. You perform an algorithm to put $M$ into Smith normal form, and the theorem is basically done. The algorithm to put a matrix in Smith normal form is recursive. You clear a column and row and then perform the algorithm again on a smaller matrix. Each time, both dimensions of the matrix decrease by one. So you can think of it as induction on $m$ if you so please.

Choosing a basis so that $mf$ is a multiple of a basis element and analyzing the torsion part is a lot like doing some steps toward Smith normal form.

  • $\begingroup$ This is helpful, but I think my approach requires less sophistication from the student. It's about a page, but written for someone who might want to teach it my way, so perhaps a little breathless for students themselves. Would you care to look and tell me if you still think you see nothing but the Smith normal form algorithm? $\endgroup$ – David Feldman Mar 29 at 21:08
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    $\begingroup$ I’ll try to take a look, where can I find it? $\endgroup$ – csprun Mar 29 at 21:56
  • $\begingroup$ Thank you. Here: we.tl/t-Qq8t14wzhC (for a week) $\endgroup$ – David Feldman Mar 29 at 22:13
  • $\begingroup$ Should I try again? $\endgroup$ – David Feldman Apr 15 at 19:49

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