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1994 Paper 3 Q7
D: 1679.5 B: 1503.1
groups permutations symmetric group S3 cyclic group isomorphism roots of unity element order modular arithmetic subgroup complex numbers

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.

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1991 Paper 2 Q9
D: 1616.2 B: 1500.0
groups matrix modular arithmetic group axioms order of group commutativity subgroup test element order upper triangular matrix

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

  • (Closure) Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are matrices of that form, then \(\begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{pmatrix}\), this is clearly of the required form since if \(a_1, a_2, c_1, c_2 \neq 0\) then \(a_1a_2, c_1c_2 \neq 0\)
  • (Associative) By inheritance from matrix multiplication
  • (Identity) Consider \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) also clearly of the required form.
  • (Inverse) Consider \((ac)^{-1}\begin{pmatrix} c & -b \\ 0 & a \end{pmatrix}\), since \(ac \neq 0\) we can assume it has an inverse mod \(5\). therefore we have another matrix of the required form.
There are \(4\) possible values for \(a\) and \(c\) and \(5\) possible values for \(b\), so \(4 \times 4 \times 5 = 80\) elements, so the group is order \(80\). \(G\) is not commutative, consider \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}\) \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix}\) The elements of order \(1\) or \(2\) satisfy \(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^{-1} & -ba^{-1}c^{-1} \\ 0 & c^{-1} \end{pmatrix}\) Therefore \(a^2 = 1, c^2 = 1 \Rightarrow a, c = 1, 4\) and \(b = -ba^{-1}c^{-1} \Rightarrow b = 0\) or , \(ac = -1\), so we have \((a,b,c) = (1,0,1), (4,0,4), (1, *, 4), (4, *, 1)\) So there are \(12\) elements of order \(1\) or \(2\). But this can't be a subgroup since \(12 \not \mid 80\)

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