Problems

Solution: \begin{questionparts} \item \((ab)^2 = abab = e\) (since \(ab\) has order \(2\)), but \(a^2 = e, b^2 = e \Rightarrow a^{-1} = a, b^{-1} = b\) (since \(a\) and \(b\) have order 2) so \(ba = ab\) by multiplication on the left by \(a\) and right by \(b\). \item Suppose \((cd)^n = e \Leftrightarrow d(cd)^nc = dc \Leftrightarrow (dc)^n(dc) = e \Leftrightarrow (dc)^n = e\) Therefore any number for which \((cd)^n = e\) has the property that \((dc)^n = e\) and vice-versa, in particular the smallest number for either \(cd\) or \(dc\) will also be the smallest number for the other. \item Given \(c^{-1}bc=b^r\), then \(b^{rs} = (b^r)^s = (c^{-1}bc)^s =\underbrace{(c^{-1}bc)(c^{-1}bc) \cdots (c^{-1}bc)}_{s \text{ times}} = c^{-1}\underbrace{bb\cdots b}_{s \text{ times}}c = c^{-1}b^sc\) We proceed by induction on \(n\). When \(n = 0\), we have \(b^s = b^{sr^0}\) so the base case is true. Suppose it is true for some \(n = k\), ie \(c^{-k}b^sc^k = b^{sr^k}\). Now consider \(c^{-{k+1}}b^sc^{k+1} = c^{-1}c^{-k}b^sc^kc = c^{-1}b^{sr^k}c = (b^{sr^k \cdot r}) = b^{sr^{k+1}}\) (where the second to last equality was by the previous part). Therefore if our statement is true for \(n=k\) it is true for \(n = k+1\). Therefore, since it is also true for \(n=0\), by the principle of mathematical induction it is true for all non-negative integers \(n\).