22 Chapter 1. Groups I a similar argument shows that (b−1a−1)(ab) = 1. The base step (ab)−1 = b−1a−1 now follows from Lemma 1.16(iii). The proof of the inductive step is left to the reader. (ii) Let every factor in part (i) be equal to a. Note that we have defined a−k = (a−1)k, and we now see that it coincides with the other worthy candidate for a−k, namely, (ak)−1. • Corollary 1.22. If G is a group, a ∈ G, and m, n ≥ 1, then am+n = aman and (am)n = amn. Proof. In the first case, both elements arise from the expression having m + n factors each equal to a in the second case, both elements arise from the expression having mn factors each equal to a. • It follows that any two powers of an element a in a group commute: aman = am+n = an+m = anam. Proposition 1.23 (Laws of Exponents). Let G be a group, let a, b ∈ G, and let m and n be (not necessarily positive) integers. (i) If a and b commute, then (ab)n = anbn. (ii) (am)n = amn. (iii) aman = am+n. Proof. The proofs, while routine, are lengthy double inductions. • The notation an is the natural way to denote a ∗ a ∗ · · · ∗ a, where a appears n times. However, if the operation is +, then it is more natural to denote a+a+···+a by na. Let G be a group written additively if a, b ∈ G and m and n are (not necessarily positive) integers, then Proposition 1.23 is usually rewritten as (i) n(a + b) = na + nb. (ii) m(na) = (mn)a. (iii) ma + na = (m + n)a. Theorem 1.20 holds in much greater generality. Definition. A semigroup is a set having an associative operation a monoid is a semigroup S having a (two-sided) identity element 1 that is, 1s = s = s1 for all s ∈ S. Of course, every group is a monoid. Example 1.24. (i) The set of natural numbers N is a commutative monoid under addition (it is also a commutative monoid under multiplication). The set of all even integers under addition is a monoid it is a semigroup under multiplication, but it is not a monoid.

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