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1999 Paper 3 Q7
D: 1680.5 B: 1516.0
groups binary operation associativity group axioms subgroup identity element inverse element

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:

  1. closure \(x*y = x + y + axy\) is a real number
  2. associativity we have already checked
  3. identity \(x*0 = x + 0 + 0 = x = 0*x\), so \(0\) is an identity
  4. inverses \(x*\left ( \frac{-x}{1+ax} \right ) = x - \frac{x}{1+ax} + ax \frac{-x}{1+ax} = \frac{x +ax^2 - x - ax^2}{1+ax} = 0\) so \(x\) has an identity, assuming \(1+ax \neq 0\) which is true for everything in our set
Consider the set \(\{0, \frac{-2}{a} \}\). Then \(\frac{-2}{a} * \frac{-2}{a} = \frac{-4}{a} + a \frac{4}{a^2} = 0\), so this is a group of order \(2\)

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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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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 2 Q10
D: 1500.0 B: 1500.0
groups group axioms binary operation closure associativity identity element inverse proof abstract algebra

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:

  1. (Closure) This operation is clearly closed by construction
  2. (Associative) \(a_j \circ (a_k \circ a_l) = a_j \circ a_{\max(k,l)} = a_{\max(j,k,l)} = a_{\max(j,k)} \circ a_l = (a_j \circ a_k) \circ a_l\), so it is associative
  3. (Identity) \(a_1 \circ a_k = a_{\max(1,k)} = a_k = a_{\max(k,1)} = a_k \circ a_1\) so \(a_1\) is an identity.
  4. (Inverses) There is no inverse, since \(a_N \circ a_k = a_N\) for all \(k\), and hence \(a_N\) can have no inverse.
  1. (Closure) \(n = |j-k|+1 \geq 1\) so we need to show that \(n \leq N\) to ensure closure. This is true since the largest \(j-k\) can be is if \(j = N\) and \(k = 1\), and this also satisfies \(|j-k| + 1 \leq N\). Hence the operation is closed.
  2. (Associative) \(a_j * (a_k * a_l) = a_j * (a_{|k-l|+1}) = a_{|j-|k-l|-1|+1}\). \((a_j * a_k) * a_l = a_{|j-k|+1}*a_l = a_{|l-|j-k|-1|+1}\). \(a_2 * (a_2 * a_3) = a_2 * a_2 = a_1\). \((a_2 *a_2)*a_3 = a_1 * a_3 = a_3 \neq a_2\) therefore this isn't associative for any \(N > 2\)
  3. (Identity) \(a_1\) is an identity, since \(a_1 * a_k = a_{|k-1|+1} = a_{k-1+1} = a_k\).
  4. (Inverse) Every element is self-inverse since \(a_k * a_k = a_{|k-k|+1} = a_1\)

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