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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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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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