Andrew Boucher's General Arithmetic (GA2) is a weak sub-theory of second order Peano Axioms (PA2). GA has second order induction and a single successor axiom:
$$\forall x \forall y \forall z\bigr((Sx=y \land Sx=z)\to(y=z)\bigl)$$
Boucher proves multiplication is commutative in GA2. Why does induction prove multiplication is commutative? GA2 has many finite models. The rings $\mathbb Z/n\mathbb Z$ are models. If we remove induction from GA2 it is easy to see GA-Ind is sub-theory of Ring Theory (RT). RT has finite non-commutative models. Why aren't these finite non-commutative rings models of GA? Would a first order version of GA also prove multiplication is commutative?
I asked on stack exchange and got no answer. http://math.stackexchange.com/questions/287557
Edit: I am not looking for an inductive proof. This is a standard result and I am sure it can be done. I am more interested in something like abo's explanation. Can we prove induction fails in every non-commutative ring? Is it impossible to define a successor chain that visits every ring element using addition in a non-commutative ring?