† The use of this terminology and the notation $\prod_{\alpha \in A} x_\alpha$ must be avoided if there is any risk of confusion with the product of the sets $x_\alpha$ defined in the theory of sets (Set Theory, II, § 5, no. 3). However, if the $x_\alpha$ are cardinals and addition (resp. multiplication) is the cardinal sum (resp. the cardinal product), the cardinal denoted by $\sum_{\alpha \in A} x_\alpha$ (resp. $\prod_{\alpha \in A} x_\alpha$) in the above notation is the cardinal sum (resp. cardinal product) of the family $(x_\alpha)_{\alpha \in A}$ (Set Theory, III, § 3, no. 3).

§ 1.3

We prove the theorem by induction on the cardinal $n$ of $A$. Let $p$ be the cardinal of $I$ and $h$ its least element; let $J = I - \{h\}$. If $n = 1$, as the $B_i$ are non-empty, of necessity $p = 1$ and the theorem is obvious. Otherwise, assuming the theorem holds for an indexing set with at most $n - 1$ elements, we distinguish two cases:

(a) $B_h$ has a single element $\beta$. Let $C = \bigcup_{i \in J} B_i$. The left-hand side of (3) is equal, by definition, to $x_\beta \top \left( \prod_{\alpha \in C} x_\alpha \right)$; the right-hand side is equal, by definition, to

the result follows from the fact that the theorem is assumed true for $C$ and $(B_i)_{i \in J}$.

(b) Otherwise, let $\beta$ be the least element of $A$ (and hence of $B_h$); let $A' = A - \{\beta\}$ and let $B'_i = A' \cap B_i$ for $i \in I$; then $B'_i = B_i$ for $i \in J$. The set $A'$ has $n - 1$ elements and the conditions of the theorem hold for $A'$ and its subsets $B'_i$; hence by hypothesis:

Forming the composition of $x_\beta$ with each side, we have on the left-hand side side, by definition, $\prod_{\alpha \in A} x_\alpha$ and on the right-hand side, using the associativity, which is equal, by Definition 3, to the right-hand side of formula (3).

For an associative law denoted by $\top$ the composition $\prod_{p \leq i \leq q} x_i$ of a sequence $(x_i)_{i \in [p, q]}$ is also denoted (since no confusion can arise) by

A particular case of Theorem 1 is the formula Consider an ordered sequence of $n$ terms all of whose terms are equal to the same element $x \in E$. The composition of this sequence is denoted by $\prod^n x$ for a law denoted by $\top$, $\prod^n x$ for a law denoted by $\perp$. For a law written multiplicatively the composition is denoted by $x^n$ and called the $n$-th power of $x$. For a law written additively the composition is usually denoted by $nx$. The associativity theorem applied to an ordered sequence all of whose terms are equal gives the equation

In particular, if $p = 2$,

and if $n_1 = n_2 = \cdots = n_p = m$,

If $X$ is a subset of $E$, we sometimes denote, in conformity with the above notation, by $\frac{p}{T} X$ the set $X_1 T X_2 T \cdots T X_p$, where

it is thus the set of all compositions $x_1 T x_2 T \cdots T x_p$ with $x_1 \in X, x_2 \in X, \ldots, x_p \in X$.

It is important not to confuse this set with the set of $\frac{p}{T} x$, where $x$ runs through $X$.

Assertions (i), (ii) and (iv) are obvious; we prove (iii). Let $\overline{X}$ be the stable subset of E generated by X and $\overline{f(X)}$ the stable subset of F generated by $f(X)$. By (i) $\overline{f(X)} \subset f(\overline{X})$ and by (ii) $\overline{X} \subset f^{-1}(\overline{f(X)})$, whence $f(\overline{X}) \subset \overline{f(X)}$.

It follows immediately by induction on n that the composition of an ordered sequence of n terms belonging to X belongs to the stable subset generated by X; it is therefore sufficient to verify that X' is stable. Now if u and v are elements of X' they are of the form $u = x_0 \top x_1 \top \cdots \top x_{n-1}$, $v = x_n \top x_{n+1} \top \cdots \top x_{n+p}$ with $x_i \in X$ for $0 \leq i \leq n + p$; then (Theorem 1) $u \top v = x_0 \top x_1 \top \cdots \top x_{n+p}$ belongs to X'.

For

Let $X'$ and $X''$ be the centralizer and bicentralizer of $X$. They are stable subsets of $E$. Now $X \subset X''$ and $Y \subset X'$ and hence $X''$ (resp. $X'$) contains the stable subset of $E$ generated by $X$ (resp. $Y$). As every element of $X''$ commutes with every element of $X'$, the proposition follows.

Since the theorem is true if A has a single element, we argue by induction on the number $p$ of elements in A. Let $p$ be an integer > 1 and suppose the theorem is true when Card A < $p$. We prove it for Card A = $p$. It may be assumed that A is the interval $[0, p - 1]$ in $\mathbf{N}$; the composition of the ordered sequence $(x_\alpha)_{\alpha \in A}$ defined by the natural order relation on A is $\prod_{i=0}^{p-1} x_i$.

Let A be given another total ordering and let h be the least element of A under this ordering and A' the set of other elements of A (totally ordered by the induced ordering). Suppose first $0 < h < p - 1$ and let P = $\{0, 1, \ldots, h - 1\}$ and Q = $\{h + 1, \ldots, p - 1\}$; the theorem being assumed true for A', applying the associativity theorem, we obtain (since $A' = P \cup Q$)

whence, composing $x_h$ with both sides and repeatedly applying the commutativity and associativity of T:

which proves the theorem for this case. If $h = 0$ or $h = p - 1$, the same result follows, but more simply, the terms arising from P or the terms arising from Q not appearing in the formulae.

Under a commutative associative law on a set E the composition of a finite family $(x_\alpha)_{\alpha \in A}$ of elements of E is by definition the common value of the composition of all the ordered sequences obtained by totally ordering A in all possible ways. This composition will still be denoted by $\prod_{\alpha \in A} x_\alpha$ under a law denoted by T; similarly for other notations.

This follows from Theorem 2 if $A$ and $I$ are totally ordered so that the $B_i$ satisfy the conditions of Theorem 1.

We single out two important special cases of this theorem:

1. If $(x_{\alpha \beta})_{(\alpha, \beta) \in A \times B}$ is a finite family of permutable elements of an associative magma whose indexing set is the product of two non-empty finite sets $A, B$ (a "double family"), then

as follows from Theorem 3 by considering $A \times B$ as the union of the sets $\{\alpha\} \times B$ on the one hand and of the sets $A \times \{\beta\}$ on the other.

For example, if $B$ has $n$ elements and for each $\alpha \in A$ all the $x_{\alpha \beta}$ have the same value $x_\alpha$, then

If $B$ has two elements, we obtain the following results: let $(x_\alpha)_{\alpha \in A}, (y_\alpha)_{\alpha \in A}$ be two non-empty families of elements of $E$. If the $x_\alpha$ and the $y_\beta$ are pairwise permutable, then

Because of formula (7) the composition of a double sequence $(x_{ij})$ whose indexing set is the product of two intervals $\{p, q\}$ and $\{r, s\}$ in $\mathbf{N}$ is often denoted for a commutative associative law written additively by

and similarly for laws denoted by other symbols.

2. Let $n$ be an integer $> 0$ and let $A$ be the set of ordered pairs of integers $(i, j)$ such that $0 \leq i \leq n, 0 \leq j \leq n$ and $i < j$; the composition of a family $(x_{ij})_{(i, j) \in A}$ (under a commutative associative law) is also denoted by $\prod_{0 \leq i < j \leq n} x_{ij}$ (or simply $\prod_{i < j} x_{ij}$, if no confusion arises); Theorem 3 here gives the formulae

There are analogous formulae to (7) for a family whose indexing set is the product of more than two sets and analogous formulae to (10) for a family whose indexing set is the set $S_p$ of strictly increasing sequences $(i_k)_{1 \leq k \leq p}$ of $p$ integers such that $0 \leq i_k \leq n$ ($p \leq n + 1$): in the latter case the composition of the family $(x_{i_1 i_2 \ldots i_p})_{(i_1, \ldots, i_p) \in S_p}$ is denoted by

For all elements x and y of E:

Theory, II, , no. 5) is a stable subset. The equivalence relations $R_A$ and $R_B$ induced by $R$ on $A$ and $B$ respectively are compatible with the induced laws and the mapping derived from the canonical injection of $A$ into $B$ by passing to the quotients is a magma isomorphism of $A / R_A$ onto $B / R_B$

Let $\top$ denote the law on $E$. If $x$ and $y$ are two elements of $B$ there exist two elements $x'$ and $y'$ of $A$ such that $x \equiv x' \pmod{R}$ and $y \equiv y' \pmod{R}$; then $x \top y \equiv x' \top y' \pmod{R}$ and $x' \top y' \in A$, whence $x \top y \in B$. Thus $B$ is a stable subset of $E$ and the other assertions are obvious.

Let $M$ be a magma and $((u_\alpha, v_\alpha))_{\alpha \in I}$ a family of elements of $M \times M$. Consider all the equivalence relations $S$ on $M$ which are compatible with the law on $M$ and such that $u_\alpha \equiv v_\alpha \pmod{S}$ for all $\alpha \in I$. The intersection of the graphs of these relations is the graph of an equivalence relation $R$ which is compatible with the law on $M$ and such that $u_\alpha \equiv v_\alpha \pmod{R}$. Hence $R$ is the finest (Set Theory, III, § 1, nos. 3 and 7) equivalence relation with these two properties. It is called the equivalence relation compatible with the law on $M$ generated by the $(u_\alpha, v_\alpha)$.

Let $T$ be the equivalence relation associated with $f$. Then $u_\alpha \equiv v_\alpha \pmod{T}$ for all $\alpha \in I$ and $T$ is compatible with the law on $M$, hence $T$ is coarser than $R$; this proves the proposition.