# Product of empty set of elements vs. product over empty indexing set — is there any difference?

I am reading Lang's *Algebra* (3rd ed., Pearson, 2003). In $\S$I.1 *Monoids*, on page 4 the author defines the meaning of and notations for products of finitely many elements of a monoid as follows:

Let $G$ be a monoid, and $x_1, \dotsc, x_n$ elements of $G$ (where $n$ is an integer $> 1$). We define their product inductively: $$ \prod_{\nu = 1}^n x_\nu = x_1 \dotsm x_n = (x_1 \dotsm x_{\nu - 1})x_\nu. $$

We then have the following rule: $$ \prod_{\mu = 1}^m x_\mu \cdot \prod_{\nu = 1}^n x_{m + \nu} = \prod_{\nu = 1}^{m + n} x_\nu, $$ which essentially asserts that we can insert parentheses in any manner in our product without changing its value.The proof is easy by induction, and we shall leave it as an exercise.One also writes $$ \prod_{m + 1}^{m + n} x_\nu \quad \text{instead of} \quad \prod_{\nu = 1}^n x_{m + \nu} $$ and we define $$ \prod_{\nu = 1}^0 x_\nu = e. $$

As a matter of convention, we agree that the empty product is equal to the unit element.

If I understand Lang correctly, in the last two points above he is trying to differentiate between the following two cases: Suppose that the elements of $G$ are indexed by elements from a linearly ordered set $I$. Then,

- if $J \subset I$, $J \neq \emptyset$ such that $\{ x_\nu \in G : \nu \in J \} = \emptyset$, then we define $\prod_{\nu \in J} x_\nu = e$;
- as a matter of convention, $\prod_{\nu \in \emptyset} x_\nu = e$.

Is there really a difference between these two cases; that is, do we need to separately define both, the product of an empty set of elements, as well as the product over an empty indexing set? Is it not possible to derive one from the other? Even better, is it not possible to derive the values of these expressions directly from the definition of products of finitely many elements itself?

One reason why Lang's exposition here doesn't feel so clean to me is that apparently many choices need to be made. I would prefer it if the definition of products of finitely many elements automatically took care of these edge cases, instead of us having to insert them in "by hand", so to speak. After all, there is only one logical choice for these values, so I expect that choice to be a consequence of the definition itself rather than it being a separate definition or convention.

## 1 answer

tl;dr: Having one logical choice for a notation that is consistent with the rest of one's definitions is not the same as *defining* that notation to mean that choice. Until one has done so, the notation has no meaning (in a strict reading of a math text).

In this excerpt, Lang is defining, presumably for convenience, two notations for the same concept of the product of a finite collection of elements from a monoid. One of those notations is $\prod_{\nu=1}^n x_\nu$, and the other notation is $x_1 \cdots x_n$. (Neither of these definitions explicitly invokes the concept of a general indexing set, just the natural numbers.)

But both notations are defined via recursion on the structure of $x_1 \cdots x_n$, and the base case of that recursion is $x_1 \cdot x_2$ (i.e., the monoid operation itself). So the empty product (the reduction of $x_1 \cdots x_n$ to the 0-element case) is undefined at this point, as is the meaning of $\prod_{\nu=1}^0 x_\nu$. If Lang plans to use either of those cases in the rest of the book, he had better say what they mean, so he does. As you note, he is constrained by the associativity rule he just laid out, so if those notations are to have any conventional meaning, they had better mean what he's defined them to mean; but he technically also has the option of simply excluding those cases from having any meaning at all, which is why one would say they're convention rather than being strict consequences of his existing definitions.

As for whether one can be derived from the other, again, we're establishing notations here, and in the initial definition of these notations Lang stipulates that $n > 1$. So the equivalence between the two notations for $n \le 1$ is not yet established.

(In fact, going off of just this excerpt (my copy of Lang is hiding in a box somewhere right now), Lang seems to leave the $n = 1$ case in an undefined limbo! You can say that *if* $\prod_{\nu=1}^1 x_\nu$ is to have a meaning consistent with the above definitions, it had better mean $x_1$, but you can't say that it *does* mean that with a strict reading of these definitions.)

I should probably also just affirm that in *informal* conversation, this distinction is rarely drawn so finely. If you were to introduce some new notation by sketching the equivalent of just the first line of math in this excerpt on a napkin or blackboard, I would expect nearly all mathematicians to fill in the blanks without you having to be explicit about it. I just wouldn't recommend it for your thesis.

#### 1 comment

I agree that this is a fairly clumsily presented definition. A precise and reasonably easy to understand definition is to define it inductively with the base case, $\prod_{\nu=1}^0 x_\nu = e$, and the inductive case: $\prod_{\nu=1}^{n+1} x_\nu = \left(\prod_{\nu=1}^n x_\nu\right)\cdot x_{n+1}$.

## 7 comments

I would have preferred to have some algebra-related tags on my question, but it seems that such tags don't already exist and I cannot create them. Could users with more privileges please help me choose a better set of tags? — Bahudari Ragam 4 months ago

Assuming $x_{(-)} : J \to G$, i.e. $x$ is a $J$-indexed collection of elements of $G$, then it can't be the case that $J \neq \varnothing$ while $\{x_\nu \in G \mid \nu \in J\} = \varnothing$. You could define a notion of product that took a finite multiset and compare that to a set-indexed notion of product to get a meaningful question. However, as r~~ points out, this is unrelated to Lang's definition which only defines a product notation for a range of naturals. — Derek Elkins 4 months ago

@DerekElkins In my mind, I was thinking of a finite monoid (or group) $G$ whose elements are indexed as $\{ x_1, \dotsc, x_n \}$ and $I = \mathbb{N}$, $J = \{ n + 1\}$. But perhaps the notion of indexing already assumes a bijection, so taking $I = \mathbb{N}$ is improper? — Bahudari Ragam 4 months ago

To be clear: am I correct to understand that by "

the last two points" you mean everything from "and we define" until the end? — Peter Taylor 3 months ago@PeterTaylor Yes, that's right. I included more than just that part in the quote because I felt that it provided context for what Lang is trying to convey; hopefully, it isn't too confusing! — Bahudari Ragam 3 months ago

Show 2 more comments