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to Artin's Algebra
Section 2.1. Laws of Composition
Exercise 2.1.1 Let \(a, b, c \in S\). Then \(a(bc) = ab = a = ac = (ab)c\) so the law of composition is associative. Obviously if \(|S| = 1\), then the sole element of \(S\) is the identity. Otherwise, suppose \(a \in S\) and choose \(b \in S \setminus \{a\}\); then \(ab = a \neq b\), so \(a\) is not the identity. Therefore there is no identity when \(|S| \neq 1\).
Exercise 2.1.2
- Suppose \(a\) has both a left inverse \(\ell\) and a right inverse \(r\), that is, \(\ell a = ar = 1\). Then by associativity, \(\ell = \ell 1 = \ell(ar) = (\ell a)r = 1r = r\). Therefore \(ar = 1\) and \(ra = \ell a = 1\), so \(r\) is the inverse of \(a\).
- If \(a\) is invertible, and \(b_1, b_2\) are inverses of \(a\), then part (a) implies that \(b_1 = b_2\). Therefore inverses are unique for associative operations.
- Observe that \((ab)(b^{-1}a^{-1}) = a(b b^{-1})a^{-1} = a a^{-1} = 1\), therefore \(b^{-1} a^{-1}\) is the inverse of \(ab\).
- If \(S\) is the set of functions \(\mathbb{N}^{\mathbb{N}}\) with the law of composition given by the composition of functions, then the function \(f \in S\) defined by \(f(x) = x+1\) has a left inverse given by \(g(x) = \max(x-1, 0)\) but no right inverse since there is no \(x\) with \(f(x) = 0\). Likewise the function \(g\) has \(f\) as its right inverse but can have no left inverse since it is not injective.
Exercise 2.1.3 This was explained in the solution to 1.2(d) above; we just note here that \(g(0)\) can be any natural number, so there are infinitely many possible \(g\)'s.