5 problems found
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) | \(\checkmark\) |
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
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 \(n\).
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: