2 problems found
The matrix \(\mathbf{M}\) is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where \(m\) is an integer greater than \(1.\) Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( and \)\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence \(\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots\) is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where \(\mathbf{P,Q}\) and \(\mathbf{X}_{0}\) are given \(2\times2\) matrices. Suggest a suitable expression for \(\mathbf{X}_{k}\) in terms of \(\mathbf{P},\) \(\mathbf{Q}\) and \(\mathbf{X}_{0},\) and justify it by induction. The binary operation \(*\) is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if \(\mathbf{P=M},\) the set \(\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}\) forms a finite group under the operation \(*\).
Solution: \(\mathbf{M}^m = \mathbf{I}\), we also have \(\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))\) therefore \(\mathbf{M-I}\) is invertible. Therefore since \(\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}\) we can cancel the \(\mathbf{M-I}\) to obtain the desired result. \(\mathbf{X_0 = X_0}\) \(\mathbf{X_1 = PX_0+Q}\) \(\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}\) Claim: \(\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}\) Proof: (By induction on \(k\)). Base case \(k = 0\) is true. Assume it's true for some \(k = l\), then consider \(k = l+1\) \(\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}\) Suppose \(\mathbf{P} = \mathbf{M}\), then consider the set \(\{\mathbf{X_1, X_2}, \ldots\}\) with the operation \(*\) as defined. \(\mathbf{X_i * X_j} = M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}\) Since \(X_m = X_0\) we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with \(m\) elements.
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 \(o(h) \neq 1\)