Associative implies generalized associative

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Let S be a set and * be a binary operation on S. Suppose that * is an associative binary operation; in other words:

a * (b * c) = (a * b) * c \ \forall \ a,b,c \in S

Then, for any positive integer n, every possible parenthesization of the expression:

a_1 * a_2 * \dots * a_n

is equivalent.


The technique of proof here is a strong form of induction: we assume the result is true for every m < n, and we prove it for n. Note that the cases n = 1,2 are tautological. For the case n = 3, there are precisely two possible parenthesizations, and associativity tells us that they are both equal.

Thus, assume n > 3, and that the result holds for all m < n. We will show that any way of parenthesization of:

a_1 * a_2 * \dots * a_n

is equal to the left-associated expression:

((\dots ( a_1 * a_2) * a_3) * \dots ) * a_n)

Let's prove this. For any method of parenthesization, there is an outermost multiplication, and in terms of this outermost multiplication, we can view the parenthesization as:

A * B

where A = a_1 * \dots * a_m and B = a_{m+1} * \dots * a_n, both parenthesized in some unknown way, with 0 < m < n. Applying the induction assumption, A equals the left-associated expression for a_1 * a_2 * \dots * a_m and B equals the left-associated expression for a_{m+1} * a_{m+2} * \dots * a_n.

Now, in the case that B is a single letter (i.e., m = n - 1), we are already done: A * B equals the left-associated expression. If not, we can write A * B as:

A * (C * a_n)

where C is the left-associated expression for a_{m+1} * \dots * a_{n-1}. Applying associativity, we get that the original expression equals:

(A * C) * a_n

which is precisely the left-associated expression for a_1 * a_2 * \dots * a_n.


  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 18-19, Section 1.1, Proposition 1 (statement on page 18, proof on page 19)