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Solution

1. We want to construct Cayley tables for $U(8)$ and $U(10)$. Then we have $$\begin{array}{c|cccc} \cdot \text{ mod } 8 & 1&3&5&7 \\ \hline\hline 1 & 1&3&5&7 \\ 3 & 3&1&7&5 \\ 5 & 5&7&1&3 \\ 7 & 7&5&3&1 \\ \end{array} \quad\quad\quad \begin{array}{c|cccc} \cdot \text{ mod } 10 & 1&7&3&9 \\ \hline\hline 1 & 1&7&3&9 \\ 7 & 7&9&1&3 \\ 3 & 3&1&9&7 \\ 9 & 9&3&7&1 \\ \end{array}.$$ Let $f:U(10)\rightarrow U(5)$ be defined by $f(x)=x\mod 5$. By inspection of the Cayley tables for $U(10)$ and $U(5)$ we see that there is a one-to-one correspondence and that the group structure is preserved. It it easy to see that when we take the elements in $U(10)$ and mod them by 5 before multiplication we will get the same result as when we take the product of the elements in $U(10)$ and mod it by 5. Thus $U(10)\approx U(5). \square$

2. In each of the Cayley tables above, we can find the identity element by finding which element when multiplied by another element returns that other element--which is the definition of the identity in a group. Let $e_{G}\in G$ and $e_H\in H$ be the identity in their respective groups. By examining the Cayley tables for $H$ and $G$ we find that $e_{G}=a$ and $e_{H}=t$. A similar proof to the one in part 1 above will show that $H\approx U(8). \square$

3. To construct the Cayley table for $K$ we have $$\begin{array}{c|cccc} \cdot & 1&-i&i&-1 \\ \hline\hline 1 & 1&-i&i&-1 \\ -i & -i&-1&1&i \\ i & i&1&-1&-i \\ -1 & -1&i&-i&1 \\ \end{array}.$$ We will show that $K\approx U(10)$. Let $f:K\rightarrow U(10)$ be defined by $\left\{ \begin{array}{l} 1\mapsto 1\\ -i\mapsto 7\\ i\mapsto 3\\ -1\mapsto 9\end{array} \right.$. Here we see that $f$ is a bijection because it is one-to-one and we can also see from the Cayley tables of $K$ and $U(10)$ that the group structure is preserved. Therefore we know $K\approx U(10). \square$

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