1 problem found
Let \(G\) be a finite group with identity \(e.\) For each element \(g\in G,\) the order of \(g\), \(o(g),\) is defined to be the smallest positive integer \(n\) for which \(g^{n}=e.\)
Solution: \begin{questionparts} \item Show that, if \(o(g)=n\) and \(g^{N}=e,\) then \(n\) divides \(N.\) Using the division algorithm, write \(N = qn + r\) where \(0 \leq r < n\) to divide \(N\) by \(n\). Then we have \(e = g^N = g^{qn + r} = g^{qn}g^r = (g^{n})^qg^r = e^qg^r = g^r\) therefore \(r\) is a number smaller than \(n\) such that \(g^r = e\). Therefore either \(r = 0\) or \(o(g) = r\), but by definition \(o(g) = n\) therefore \(r = 0\) and \(n \mid N\). \item Let \(g\) and \(h\) be elements of \(G\). Prove that, for any integer \(m,\) \[ gh^{m}g^{-1}=(ghg^{-1})^{m}. \] \((ghg^{-1})^m = \underbrace{(ghg^{-1})(ghg^{-1})\cdots(ghg^{-1})}_{m \text{ times}} = gh(g^{-1}g)h(g^{-1}g)\cdots (g^{-1}g)hg^{-1} = gh^mg^{-1}\) \item Let \(g\) and \(h\) be elements of \(G\), such that \(g^{5}=e,h\neq e\) and \(ghg^{-1}=h^{2}.\) Prove that \(g^{2}hg^{-2}=h^{4}\) and find \(o(h).\) \(g^2hg^{-2} = g(ghg^{-1})g^{-1} = gh^2g^{-1} = (ghg^{-1})^2 = (h^2)^2 = h^4\). \(h = g^{5}hg^{-5} = g^4ghg^{-1}g^{-4} = g^4h^2g^{-4} = g^3(ghg^{-1})^2g^{-3} = g^3h^4g^{-3} = h^32\) Therefore \(e = h^{31}\). Therfore \(o(h) \mid 31 \Rightarrow \boxed{o(h) = 31}\) since \(31\) is prime and \(o(h) \neq 1\)