14 problems found
Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.
Solution: Claim: \(*\) is associative. Proof: Then \(x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz\) and \((x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz\) so \(x*(y*z) = (x*y)*z\) and we are done. Let \(G = \mathbb{R} \setminus \{-\frac1{a} \}\) In order to show that \((G, *)\) is a group we need to check:
Solution:
Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.
Solution:
| 1 | 3 | 5 | 7 | |
| 1 | 1 | 3 | 5 | 7 |
| 3 | 3 | 1 | 7 | 5 |
| 5 | 5 | 7 | 1 | 3 |
| 7 | 7 | 5 | 3 | 1 |
| 1 | 2 | 3 | 4 | |
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | 4 | 1 | 3 |
| 3 | 3 | 1 | 4 | 2 |
| 4 | 4 | 3 | 2 | 1 |
| (i) | (iii) | (iv) | (vi) | |
| (i) | \(\checkmark\) | \(\checkmark\) consider \(z \mapsto \begin{pmatrix} \cos \arg (z) | - \sin \arg(z) | |
| \sin \arg(z) | \cos \arg(z) \end{pmatrix}\) | not finite | not finite | |
| (iii) | \(\checkmark\) | not finite | not finite | |
| (iv) | \(\checkmark\) | no element order \(4\) | ||
| (vi) | \(\checkmark\) |
Let \(S_{3}\) be the group of permutations of three objects and \(Z_{6}\) be the group of integers under addition modulo 6. List all the elements of each group, stating the order of each element. State, with reasons, whether \(S_{3}\) is isomorphic with \(Z_{6}.\) Let \(C_{6}\) be the group of 6th roots of unity. That is, \(C_{6}=\{1,\alpha,\alpha^{2},\alpha^{3},\alpha^{4},\alpha^{5}\}\) where \(\alpha=\mathrm{e}^{\mathrm{i}\pi/3}\) and the group operation is complex multiplication. Prove that \(C_{6}\) is isomorphic with \(Z_{6}.\) Is there any (multiplicative or additive) subgroup of the complex numbers which is isomorphic with \(S_{3}\)? Give a reason for your answer.
Solution: \(S_3 \) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & e & (12) & (13) & (23) & (123) & (132) \\ \text{order} & 1 & 2 & 2 & 2 & 3 & 3 \\ \end{array}$ \(\mathbb{Z}_6\) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & 0 & 1 & 2 & 3 & 4 & 5 \\ \text{order} & 1 & 6 & 3 & 2 & 3 & 6 \\ \end{array}$ \(S_3\) is not isomorphic to \(\mathbb{Z}_6\) since \(\mathbb{Z}_6\) has two elements of order \(6\) but \(S_3\) has none. Consider the map \(f : \mathbb{Z}_6 \to C_6\) with \(i \mapsto \alpha^i\). This is an isomorphism, since \(i + j \mapsto \alpha^{i+j} = \alpha^i\alpha^j\) \(S_3\) is non-abelian, since \((12)(123) = (23) \neq (13) = (123)(12)\) but multiplication and addition of complex numbers is commutative.
The set \(S\) consists of ordered pairs of complex numbers \((z_1,z_2)\) and a binary operation \(\circ\) on \(S\) is defined by $$ (z_1,z_2)\circ(w_1,w_2)= (z_1w_1-z_2w^*_2, \; z_1w_2+z_2w^*_1). $$ Show that the operation \(\circ\) is associative and determine whether it is commutative. Evaluate \((z,0)\circ(w,0)\), \((z,0)\circ(0,w)\), \((0,z)\circ(w,0)\) and \((0,z)\circ(0,w)\). The set \(S_1\) is the subset of \(S\) consisting of \(A\), \(B\), \(\ldots\,\), \(H\), where \(A=(1,0)\), \(B=(0,1)\), \(C=(i,0)\), \(D=(0,i)\), \(E=(-1,0)\), \(F=(0,-1)\), \(G=(-i,0)\) and \(H=(0,-i)\). Show that \(S_1\) is closed under \(\circ\) and that it has an identity element. Determine the inverse and order of each element of \(S_1\). Show that \(S_1\) is a group under \(\circ\). \hfil\break [You are not required to compute the multiplication table in full.] Show that \(\{A,B,E,F\}\) is a subgroup of \(S_1\) and determine whether it is isomorphic to the group generated by the \(2\times2\) matrix $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ under matrix multiplication.
Explain what is meant by the order of an element \(g\) of a group \(G\). The set \(S\) consists of all \(2\times2\) matrices whose determinant is \(1\). Find the inverse of the element \(\mathbf{A}\) of \(S\), where \[ \mathbf{A}=\begin{pmatrix}w & x\\ y & z \end{pmatrix}. \] Show that \(S\) is a group under matrix multiplication (you may assume that matrix multiplication is associative). For which elements \(\mathbf{A}\) is \(\mathbf{A}^{-1}=\mathbf{A}\)? Which element or elements have order 2? Show that the element \(\mathbf{A}\) of \(S\) has order 3 if, and only if, \(w+z+1=0.\) Write down one such element.
Solution: The order of an element \(g\) is the smallest positive number \(k\) such that \(g^k = e\). $\mathbf{A}^{-1} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Claim, \(S\) is a group. \begin{enumerate} \item (Closure) The product of two \(2\times2\) matrices is always a \(2\times 2\) matrix so we only need to check the determinant. Suppose \(\det(\mathbf{A}) = \det (\mathbf{B}) = 1\), then \(\det(AB) = \det(A)\det(B) = 1\), so our operation is closed \item (Associativity) Inherited from matrix multiplication \item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}\( has determinant \)1$. \item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer. \end{itemize} If \(\mathbf{A}^-1 = \mathbf{A}\) then assuming $\mathbf{A} = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\( then \)\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\ -c & a \end{pmatrix}\( so we must have \)a=d, -b=b, -c=c\(, so \)b = c = 0\( and \)a = d\(. For the determinant to be \)1\( we must have \)ad = a^2 = 1\(, ie \)a = \pm 1\(. Therefore we must have \)\mathbf{A} = \begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( or \)\mathbf{A} = \begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(2\) then \(\mathbf{A}^2 = \mathbf{I}\) ie, \(\mathbf{A} = \mathbf{A}^{-1}\) and \(\mathbf{A} \neq \mathbf{I}\) therefore the only element of order \(2\) is $\begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(3\) we must have \(\mathbf{A}^2 = \mathbf{A}^{-1}\), ie $\begin{pmatrix}w^2 + xy & x(w+z)\\ y(w+z) & z^2 + xy \end{pmatrix} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Therefore \(w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w\). The second and third equations are satisfied iff \(w+z+1 = 0\) or \(x = 0\) and \(y = 0\), but if \(x = 0\) and \(y = 0\) then we aren't order \(3\), so we just need to check this is sufficient for the first and last equations. Since \(\det(\mathbf{A}) = 1\) we have \(wz =xy +1\), so the first and last equations are equivalent to \(w^2 + wz - 1 = z\) and \(x^2 + wz-1 = w\) which are equivalent to \(w(w+z) = z+1\) or \(w + z+ 1 = 0\) as required
Let \(G\) be the set of all matrices of the form \[ \begin{pmatrix}a & b\\ 0 & c \end{pmatrix}, \] where \(a,b\) and \(c\) are integers modulo 5, and \(a\neq0\neq c\). Show that \(G\) forms a group under matrix multiplication (which may be assumed to be associative). What is the order of \(G\)? Determine whether or not \(G\) is commutative. Determine whether or not the set consisting of all elements in \(G\) of order \(1\) or \(2\) is a subgroup of \(G\).
Solution: Claim \(G\) is a group under matrix multiplication
The elements \(a,b,c,d\) belong to the group \(G\) with binary operation \(*.\) Show that
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 \(n\).
Let \(\Omega=\exp(\mathrm{i}\pi/3).\) Prove that \(\Omega^{2}-\Omega+1=0.\) Two transformations, \(R\) and \(T\), of the complex plane are defined by \[ R:z\longmapsto\Omega^{2}z\qquad\mbox{ and }\qquad T:z\longmapsto\dfrac{\Omega z+\Omega^{2}}{2\Omega^{2}z+1}. \] Verify that each of \(R\) and \(T\) permute the four point \(z_{0}=0,\) \(z_{1}=1,\) \(z_{2}=\Omega^{2}\) and \(z_{3}=-\Omega.\) Explain, without explicitly producing a group multiplication table, why the smallest group of transformations which contains elements \(R\) and \(T\) has order at least 12. Are there any permutations of these points which cannot be produced by repeated combinations of \(R\) and \(T\)?
Solution: \(R(0) = 0\), \(R(1) = \Omega^2 1 = \Omega^2\), \(R(\Omega^2) = \Omega^4 = -\Omega\), \(R(-\Omega) = -\Omega^3 = 1\) \(T(0) = \frac{\Omega^2}1 = \Omega^2\), \(T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1\) \(T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega\) \(T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0\) Thinking of \(R\) and \(S\) as elements of \(S_4\), we have that \(R = (234), S = (134)\), we can also construct \(RS = (14)(23), R^2S = (12)(34), RSR^2S = (13)(24)\). Therefore we have the subgroups \(\{e, (234), (243)\}\) of order \(3\) and the subgroup \(\{e, (12)(34), (13)(24), (14)(23) \}\) of order \(4\). By Lagrange's theorem this means that both \(3\) and \(4\) divide the order of the group, therefore the group has order divisible by \(12\) (and therefore is at least \(12\)). Yes, we cannot produce any odd permutation, for example \((12)\) cannot be produced. (Since all our generators are even permutations).
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.
Give a careful argument to show that, if \(G_{1}\) and \(G_{2}\) are subgroups of a finite group \(G\) such that every element of \(G\) is either in \(G_{1}\) or in \(G_{2},\) then either \(G_{1}=G\) or \(G_{2}=G\). Give an example of a group \(H\) which has three subgroups \(H_{1},H_{2}\) and \(H_{3}\) such that every element of \(H\) is either in \(H_{1},H_{2}\) or \(H_{3}\) and \(H_{1}\neq H,H_{2}\neq H,H_{3}\neq H\).
Solution: Suppose \(|G_1|, |G_2| < |G|\) for sake of contraction. Then by Lagrange's theorem \(|G_1| \mid |G|\) and \(|G_2| \mid |G|\), so \(|G_1|, |G_2| \leq \frac{|G|}{2}\). But \(|G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|\). \(|G_1 \cup G_2| = |G|\) by assumption, and \(e \in G_1 \cap G_2\), so \(|G_1 \cap G_2| \geq 1\). Therefore \(|G| = |G_1| + |G_2| - |G_1 \cap G_2| \leq \frac{|G|}{2} + \frac{|G|}{2} - 1 = |G| - 1 < |G|\), contradiction! Let \(H = K_4 = \{e, a, b, c\}\) with \(a^2 = b^2 = c^2 = e\). then \(H = \{e, a\} \cup \{e, b\} \cup \{e, c\}\) with all subgroups distinct
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 \(o(h) \neq 1\)
The set \(S\) consists of \(N(>2)\) elements \(a_{1},a_{2},\ldots,a_{N}.\) \(S\) is acted upon by a binary operation \(\circ,\) defined by \[ a_{j}\circ a_{k}=a_{m}, \] where \(m\) is equal to the greater of \(j\) and \(k\). Determine, giving reasons, which of the four group axioms hold for \(S\) under \(\circ,\) and which do not. Determine also, giving reasons, which of the group axioms hold for \(S\) under \(*\), where \(*\) is defined by \[ a_{j}*a_{k}=a_{n}, \] where \(n=\left|j-k\right|+1\).
Solution:
Let \((G,*)\) and \((H,\circ)\) be two groups and \(G\times H\) be the set of ordered pairs \((g,h)\) with \(g\in G\) and \(h\in H.\) A multiplication on \(G\times H\) is defined by \[ (g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2}) \] for all \(g_{1},g_{2}\in G\) and \(h_{1},h_{2}\in H\). Show that, with this multiplication, \(G\times H\) is a group. State whether the following are true or false and prove your answers.
Solution: Claim: \(G \times H\) is a group. (Called the product group). Proof: Checking the group axioms: