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1995 Paper 3 Q7
D: 1654.7 B: 1516.0
groups group axioms closure identity element inverse isomorphism complex numbers modular arithmetic matrix rotation matrix unit modulus

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.

  1. the complex numbers of unit modulus;
  2. the integers modulo 4;
  3. the matrices \[ \mathrm{M}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}, \] where \(0\leqslant\theta<2\pi\);
  4. the integers \(1,3,5,7\) modulo 8;
  5. the \(2\times2\) matrices all of whose entries are integers;
  6. the integers \(1,2,3,4\) modulo 5.
In the case of each pair of groups above state, with reasons, whether or not they are isomorphic.

Solution:

  1. \(\{ z \in \mathbb{C} : |z| = 1\}\) is a group.
    1. (Closure) \(|z_1z_2| = |z_1||z_2| = 1\). Set is closed under multiplication
    2. (Associativity) Multiplication of complex numbers is associative
    3. (Identity) \(|1| = 1\)
    4. (Inverses) \(| \frac{1}{z} | = \frac{1}{|z|} = \frac{1}{1} = 1\), the set contains inverses
  2. the integers \(\pmod{4}\) are not a group under multiplication, \(2\) has no inverse, since \(0 \times k \equiv 0 \pmod{4}\)
  3. The set of rotation matrices is a group:
    1. (Closure) \begin{align*} \begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1 \end{pmatrix} \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2 \end{pmatrix} &= {\scriptscriptstyle\begin{pmatrix}\cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 & -\sin\theta_1\ \cos \theta_1 - \sin \theta_2\cos\theta_1\\ \sin\theta_1\ \cos \theta_1 + \sin \theta_2\cos\theta_1 & \cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 \end{pmatrix}} \\ &= \begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2) \end{pmatrix} \end{align*} Since \(\cos, \sin\) are periodic with period \(2\pi\), we can find \(\theta_3 = \theta_1 + \theta_2 + 2k\pi\) such that \(0 \leq \theta_3 < 2 \pi\), so our set is closed
    2. (Associativity) Matrix multiplication is associative
    3. (Identity) Consider \(\theta = 0\)
    4. (Inverses) Consider \(2\pi - \theta\)
  4. \(\{1, 3, 5, 7\} \pmod{8}\) is a group:
    1357
    11357
    33175
    55713
    77531
    1. (Closure) See Cayley table
    2. (Associativity) Integer multiplication is associative
    3. (Identity) \(1\)
    4. (Inverses) \(x \mapsto x\) (See Cayley table)
  5. \(2\times2\) matrices are not a group, consider $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\(, then \)\mathbf{0}\mathbf{M} = \mathbf{0}$ for all other matrices.
  6. 1234
    11234
    22413
    33142
    44321
    1. (Closure) See Cayley table
    2. (Associativity) Integer multiplication is associative
    3. (Identity) \(1\)
    4. (Inverses) \(1 \mapsto 1, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 4\) (See Cayley table)
(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 finitenot finite
(iii)\(\checkmark\)not finitenot finite
(iv)\(\checkmark\)no element order \(4\)
(vi)\(\checkmark\)

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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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1993 Paper 3 Q5
D: 1730.5 B: 1466.6
groups binary operation associativity commutativity complex numbers subgroup isomorphism order of elements identity element inverse matrix quaternion group

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.

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1987 Paper 3 Q9
D: 1500.0 B: 1500.0
groups direct product abelian group subgroup isomorphism cyclic group permutation group group axioms proof

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.

  1. \(G\times H\) is abelian if and only if both \(G\) and \(H\) are abelian.
  2. \(G\times H\) contains a subgroup isomorphic to \(G\).
  3. \(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\) is isomorphic to \(\mathbb{Z}_{4}.\)
  4. \(S_{2}\times S_{3}\) is isomorphic to \(S_{6}.\)
{[}\(\mathbb{Z}_{n}\) is the cyclic group of order \(n\), and \(S_{n}\) is the permutation group on \(n\) objects.{]}

Solution: Claim: \(G \times H\) is a group. (Called the product group). Proof: Checking the group axioms:

  1. (Closure) is inherited from \(G\) and \(H\), since \(g_1 * g_2 \in G\) and \(h_1 \circ h_2 \in H\)
  2. (Associativity) \begin{align*} (g_1, h_1)\l (g_2, h_2)(g_3,h_3)\r &= (g_1, h_1)(g_2 *g_3, h_2 \circ h_3) \\ &= (g_1*(g_2 *g_3), h_1 \circ (h_2 \circ h_3)) \\ &= ((g_1*g_2) *g_3), (h_1 \circ h_2) \circ h_3) \\ &= (g_1*g_2, h_1 \circ h_2)(g_3, h_3) \\ &= \l(g_1, h_1)(g_2, h_2) \r(g_3,h_3) \end{align*}
  3. (Identity) Consider \((e_G, e_H)\), then \((e_G, e_H)(g,h) = (g,h) = (g,h)(e_G, e_H)\)
  4. (Inverses) If \((g,h) \in G \times H\) then consider \((g^{-1}, h^{-1})\) and we have \((g^{-1}, h^{-1})(g,h) = (e_G,e_H) = (g,h)(g^{-1}, h^{-1})\)
  • Claim: \(G \times H\) is abelian iff \(G\) and \(H\) are. Proof: \(\Rightarrow\) Suppose \(g_1, g_2 \in G\) and \(h_1, h_2 \in H\) then \((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (h_1,h_2)(g_1, g_2) = (g_2 * g_1, h_2 \circ h_1)\) so \(g_1*g_2 = g_2*g_1\) and \(h_1 \circ h_2 = h_2 \circ h_1\), therefore \(G\) and \(H\) are commutative. \(\Leftarrow\) If \(H\) and \(G\) are commutative then: \((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (g_2 * g_1, h_2 \circ h_1) = (h_1,h_2)(g_1, g_2)\) so \(G \times H\) is commutative.
  • Claim: \(G\times H\) contains a subgroup isomorphic to \(G\). Consider the subset \(S = \{(g,e_H) : g \in G \}\). Then this is a subgroup isomorphic to \(G\) with isomorphism given by \(\phi : S \to G\) by \(\phi((g,e_H)) = g\)
  • If \(x \in \mathbb{Z}_2 \times \mathbb{Z}_2\) then \(x^2 = e\), but \(1\) does not have order 2 in \(\mathbb{Z}_4\)
  • \(S_2 \times S_3\) has order \(2 \times 6 = 12\). \(S_6\) has order \(6! \neq 12\)

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