1. (i) $\rho^{-1} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 1 & 4 & 2 & 5 & 3\end{pmatrix}$ $\rho \tau = \begin{pmatrix}1 & 2 & 3 &4 &5 \\ 5 & 1 & 4 & 3 & 2 \end{pmatrix}$ $\tau^{2}= \begin{pmatrix}1 & 2& 3& 4& 5 \\ 5 & 3 & 4 & 1 & 2 \end{pmatrix}$ (ii) Repeatedly applying $\rho$ to $1$ gives $1 \mapsto 1$Repeatedly applying $\rho$ to $2$ gives $2 \mapsto 3 \mapsto 5 \mapsto 4 \mapsto 2$Therefore $\rho = (1)(2\;3\;5\;4)$ (iii) $\rho$ is odd, $\tau = (1 \; 3\; 5\; 4\; 2)$ which is even. 2. (i) 2 (ii) 3 (iii) 6 (iv) $(1\;2\;3)(1\;2) = (1 \;3)$ which has order 2. 3. (a) Let $(G, \diamond)$ and $(H,*)$ be isomorphic groups. Hence there exists a bijection $\phi: G \to H$ such that $\phi(g_{1} \diamond g_{2}) = \phi(g_{1}) * \phi(g_{2}) \quad \forall g_{1},g_{2} \in G$Suppose there exists $g \in G$ that has order $n$. We can write $g^{n}= 1_{G}$, the identity of $G$. Applying $\phi$ to both sides gives $\phi(g^{n}) = \phi(1_{G}) \implies \phi(g)^{n} =1_{H}$, the identity of $H$. Now suppose there exists $0<m<n$ such that $\phi(g)^{m} = 1_{H}$. Deduce that $\phi(g^{m})=\phi(1_{G})$ Since $\phi$ is injective, $g^{m} = 1_{G}$ which contradicts our assumption that $g$ has order $n$. So $\phi(g)$ must have order $n$. (b) $[1]_{6} \in \mathbb{Z} \text{/} 6 \mathbb{Z}$ has order 6. However no element of $D_{6}$ has order 6. Which shows that $(\mathbb{Z} \text{/} 6\mathbb{Z}, +_{6})$ and $(D_{6}, \circ)$ are not isomorphic. (c)The elements of $\mathbb{Z} \text{/}2\mathbb{Z} \times \mathbb{Z} \text{/}4\mathbb{Z}$ are of the form $(a,b)$ where $a \in \mathbb{Z} \text{/}2\mathbb{Z}, b \in \mathbb{Z} \text{/}4\mathbb{Z}$ There are $2$ choices for $a$ and $4$ choices for $b$. Hence their direct product has order $2 \times 4=8$. (d) $\mathbb{Z} \text{/}2\mathbb{Z} \times \mathbb{Z} \text{/}4\mathbb{Z}$ is abelian whereas $D_{8}$ is non-abelian so they cannot be isomorphic. 4. Let $G$ be homomorphic to $H$. Hence there exists a function $\phi: G \to H$ such that $\phi(g_{1}g_{2}) = \phi(g_{1})\phi(g_{2}) \quad \forall g_{1}, g_{2} \in G$. (a) Suppose that $G$ abelian and $\phi$ is surjective. Since $\phi$ is surjective, for all $h_{1}, h_{2} \in H$, there exists $g_{1}, g_{2} \in G$ such that $h_{1} = \phi(g_{1})$ and $h_{2} = \phi(g_{2})$. So $\begin{align}h_{1}h_{2} & = \phi(g_{1}) \phi(g_{2}) \\ &= \phi(g_{1} g_{2}) & \text{(by definition of a homomorphism) } \\ &= \phi(g_{2} g_{1}) &\text{(since G is abelian)} \\ &= \phi(g_{2}) \phi(g_{1} ) \\ &= h_{2} h_{1} \end{align}$which shows that $H$ is also abelian. (b) Now suppose $H$ is abelian and $\phi$ injective. For all $g_{1},g_{2} \in G$, $\begin{align}\phi(g_{1}g_{2}) &= \phi(g_{1})\phi(g_{2}) & \text{(by definition of a homomorphism)} \\ &=\phi(g_{2})\phi(g_{1}) & \text{(since H is abelian and } \phi(g_{1}), \phi(g_{2} ) \in H) \\ &= \phi(g_{2}g_{1})\end{align}$Since $\phi$ is injective, $g_{1}g_{2} = g_{2} g_{1}$ which shows that $G$ is abelian. (c) As a result of (a), there is no surjective homomorphism from an abelian group to a non-abelian group. However we can define a non-surjective homomorphism from an abelian group to a non-abelian group. For example, let $H = D_{6}$ and $G$ be the subgroup $\{ \rho_{0}, \rho_{1}, \rho_{2} \}$. Define $\phi: G \to H$ as $\phi(m) = m$. For all $m, n \in G$, $\phi(mn)=mn=\phi(m)\phi(n)$ hence $\phi$ is a homomorphism. Clearly $G$ is abelian. However $H$ is non-abelian. 5. Let $G$ be a group and fix $g \in G$ Define the function $\phi: G \to G$ by $\phi(h) = g^{-1} hg$ for all $h \in G$ Now $\begin{align}\phi(h_{1}h_{2}) &= g^{-1} (h_{1} h_{2}) g \\ &= (g^{-1} h_{1})(1) (h_{2} g) &\text{(using associativity and identity 1)}\\ &= (g^{-1} h_{1})(g g^{-1} ) (h_{2} g) \\ &= (g^{-1} h_{1}g) (g^{-1} h_{2} g) \\ &= \phi(h_{1})\phi(h_{2})\end{align}$ Take $h_{1},h_{2} \in G$ such that $\phi(h_{1}) =\phi(h_{2})$ Then $g^{-1} h_{1} g = g^{-1} h_{2} g$. Pre- and post- multiplying both sides by $g$ and $g^{-1}$ respectively gives: $\begin{gather} g(g^{-1}h_{1} g)g^{-1} = g(g^{-1}h_{2} g)g^{-1} \\ \implies (gg^{-1})h_{1} (gg^{-1}) = (gg^{-1})h_{2} (gg^{-1}) &\text{(using associativity)}\\ \implies h_{1} = h_{2}\end{gather}$Which shows that $\phi$ is injective. Now take any $h \in G$. Using associativity, $\phi(ghg^{-1)}= g^{-1} (g h g^{-1}) g= (g^{-1} g) h (g^{-1} g) = h$. Also $ghg^{-1} \in G$ as a result of closure. Hence $\phi$ is surjective. Therefore $\phi$ is bijective and for all $h_{1},h_{2} \in G$, $\phi(h_{1}h_{2}) = \phi(h_{1})\phi(h_{2})$ so $\phi$ is an isomorphism from $G$ to itself.