The product $g_0{g}_1\cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n=0$) equal to the identity

Let $(M,\cdot ,e)$ be a monoid (Semigroup and monoid) and let $g:\mathbb{N}\to M$ be a family of elements of $M$, written $g_i:=g(i)$. There is exactly one function $P_g:\mathbb{N}\to M$ satisfying

$P_g(0)=e,P_g(\sigma (n))=P_g(n)\cdot g_n(n\in \mathbb{N}),$

and we write

${\prod }_{i<n}{g}_i:=P_g(n),\text{also written }{g}_0{g}_1\cdots g_{n-1}\mathrm{.}$

In particular the empty product is ${\prod }_{i<0}{g}_i=e$, and ${\prod }_{i<1}{g}_i=e\cdot g_0=g_0$.

Why the recursion is legitimate. The clause $P_g(\sigma (n))=P_g(n)\cdot g_n$ consults $n$ as well as $P_g(n)$, so The recursion theorem does not apply to it directly. Apply that theorem instead with the set $A=\mathbb{N}\times M$, the element $a=(0,e)$, and the function $F:A\to A$ given by $F(n,x)=(\sigma (n),x\cdot g_n)$: it yields a unique $H:\mathbb{N}\to \mathbb{N}\times M$ with $H(0)=(0,e)$ and $H(\sigma (n))=F(H(n))$. Writing $H(n)=(H_1(n),H_2(n))$, induction (The principle of mathematical induction) gives $H_1(n)=n$ for every $n$, since $H_1(0)=0$ and $H_1(\sigma (n))=\sigma (H_1(n))$. Hence $H(\sigma (n))=(\sigma (n),H_2(n)\cdot g_n)$, so $P_g:=H_2$ satisfies the two displayed equations. It is the only such function: if $Q$ satisfies them too, then $\{n:P_g(n)=Q(n)\}$ contains $0$ and is closed under $\sigma$, hence is all of $\mathbb{N}$ by induction.

The value depends only on $g_0,\dots ,g_{n-1}$. If $g,g^{\mathrm{\prime }}:\mathbb{N}\to M$ satisfy $g_i=g_i^{\mathrm{\prime }}$ for every $i<n$, then $P_g(n)=P_{g^{\mathrm{\prime }}}(n)$. Indeed the set of $n$ for which this implication holds contains $0$, both products then being $e$; and if it holds at $n$, and $g,g^{\mathrm{\prime }}$ agree at every $i<\sigma (n)$, then they agree at every $i<n$ and also at $n$ itself, because $i<\sigma (n)$ is equivalent to $i\le n$ (On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), so $P_g(\sigma (n))=P_g(n)\cdot g_n=P_{g^{\mathrm{\prime }}}(n)\cdot g_n^{\mathrm{\prime }}=P_{g^{\mathrm{\prime }}}(\sigma (n))$. Induction finishes it. This is what makes the notation $g_0{g}_1\cdots g_{n-1}$ unambiguous: it names a value determined by the first $n$ terms alone, and a finite list $u$ of length $n$, that is a function $u:n\to M$ on the von Neumann natural $n=\{0,\dots ,n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann)), determines the product ${\prod }_{i<n}{u}_i:=P_{\tilde{u}}(n)$ computed from any extension $\tilde{u}:\mathbb{N}\to M$ of $u$.

• The empty product is the identity, and this is not a convention chosen for convenience. It is forced by the recursion, whose base clause is $P_g(0)=e$, and it is what makes the induction in Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either start. $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)), so $n=0$ is a genuine case of every statement below, never an afterthought.

• The factors are multiplied left to right: $P_g(\sigma (n))$ appends $g_n$ on the right. Nothing depends on that choice, because Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either shows the value is unchanged by any regrouping of consecutive factors; but the choice must be made, since the recursion has to say where the new factor goes.

• The existing Finite sums and finite products, by recursion is a different object: it is stated for sequences $\mathbb{N}\to \mathbb{R}$ in the complete ordered field, and its multiplicative alias already names the product of real numbers. It cannot carry a product in an arbitrary monoid, which is why this item exists.