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In several abstract algebra textbooks, I have been seeing propositions that I would think require induction verified without using induction. For example, consider the claim that if $G_{1}, \ldots, G_{n}$ is a collection of groups, then the direct product $G_{1} \times \cdots \times G_{n}$ is a group. My initial reaction is to prove that the claim holds for $n=2$, specifically $G \times H$ is a group if $G$ and $H$ are groups, to establish the basis step, then to proceed with the induction step, and finally to conclude, by the principle of mathematical induction, that our desired result is true. However, in the textbooks, they simply let $G_{1}, \ldots, G_{n}$ be a collection of groups, show that $(e_{G_{1}},\ldots, e_{G_{n}})$ is the identity and so on and so forth -- without explicitly mentioning induction. Do these proof not require induction? I am slightly confused.

As another example, I posted this solution for verification earlier:

Center of Direct Product is the Direct Product of Centers

Initially, I used induction-as you can see. However, I also noticed that it is more easily proved by jumping right in-see my question at the bottom. I received excellent feedback, which confirmed my induction, but also suggested the easier way (i.e. the way these claims are being proved in the texts). Can someone please help clear up my confusion? Thank you so much!

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dgc1240
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  • Induction is used, but the proof follows a very easy path. – egreg Jan 07 '14 at 22:22
  • I'm not sure that I understand your confusion -- there are a lot of results that can be proved in a variety of different ways. A claim can perfectly well have an inductive proof and another proof. What you prefer is really a matter of your background, taste, and the moment! – Nick Peterson Jan 07 '14 at 22:28
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    Strictly speaking any time, you write an expression with and ellipsis ("$\ldots$") you are invoking induction since you are claiming that the ellipsis could stand for $0$, $1$, $2$, etc. elements in a prescribed pattern. – Sammy Black Jan 07 '14 at 22:34
  • I think, whenever we use $\Bbb N$, induction must be there on platform level. But once we suppose that $\Bbb N$ is given, no more induction is needed for $G_1\times\dots\times G_n$... (in my feeling) – Berci Jan 07 '14 at 22:41
  • I think what happens is that after you get to a certain point it's assumed that you recognize that any proof with a $\dots$ in it is really an induction proof, and that you are capable of supplying the details on your own. – Gerry Myerson Jan 07 '14 at 22:57

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I think the source of confusion is the unrigorous use of the phrase "a collection of things". A better phrase than "collection of things" is "family of things indexed by something else" or "parametrized family" since what we really care about is a collection of things unified by being parametrized/indexed by a common third thing.

In particular, we can move the ellipses in the statement of the question a bit, so that it says the following: "If we have a family of groups $G_i$ indexed by (the set) $I_n=\{1,2,\dots,n\}$, then the Cartesian product of this family of groups is also a group (or rather, has a natural group structure)".

Then if we think carefully about the proof of the above statement, we will see that it is really the proof of the statement "If we have a family of groups $G_i$ indexed by (a set) $I$, then the Cartesian product of this family of groups is also a group". Certainly, any notion of induction for the natural numbers here is irrelevant: the only relevant notions are the ones for manipulating families of objects indexed by a set (these are probably left as unrigorous in your book as set theory is, since the modest requirements of algebra do not lead to the potential contradiction that require careful axiomatic formulation).

On the other hand, the ellipsis in $I_n=\{1,2,\dots,n\}$ indicate that this is a set defined by recursion, i.e. that $I_{\cdot}$ is a recursive function with domain the natural numbers. Hence, for the statement of the problem to even make sense, it is necessary to use recursion, since otherwise the index set which indexes the family of groups will be undefined. From this point onward, one could either prove the statement using the intuitive rules for manipulating families over arbitrary sets (which depend on comprehension and replacement), or one could use recursion and induction to make rigorous the notion of a family over a subset of $\mathbb N$, and then either prove all the rules for manipulating such families, or just the rules needed to carry out the proof of the statement.

Vladimir Sotirov
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