1) We know that \(\displaystyle{\mathbb{Z}_{{n}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{e}{r}{e}{\gcd{{\left({a},{n}\right)}}}={1},{1}\le{a}\le{n}\right\rbrace}\) form a froup under multiplication with inverse bar1. Now,

\(\displaystyle{\mathbb{Z}_{{5}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{a}{e}{r}{e}{\gcd{{\left({a},{5}\right)}}}={1},{1}\le{a}\le{5}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{2}},\overline{{3}},\overline{{4}}\right\rbrace}.\)

Check that

\(\displaystyle{\left(\overline{{2}}\right)}^{{4}}={\left(\overline{{4}}\right)}^{{2}}={1},\)

\(\displaystyle{\left({4}^{{2}}={16}\equiv\in\text{mod}{5}\right)}\)

As here 4 is the element positive integer such that \(\displaystyle\overline{{2}}^{{4}}=\overline{{1}}\), therefore, order of \(\displaystyle\overline{{2}}\ne{\mathbb{Z}_{{5}}^{\ast}}={4}\). Hence, \(\displaystyle{\mathbb{Z}_{{5}}^{\ast}}\) is cyclic

2) Again \(\displaystyle{\mathbb{Z}_{{8}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{a}{e}{r}{e}{\gcd{{\left({a},{8}\right)}}}={1},{1}\le{a}\le{8}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{3}},\overline{{5}},\overline{{7}}\right\rbrace}.,\)

but \(\displaystyle\overline{{3}}^{{2}}=\overline{^}{2}=\overline{{7}}^{{2}}=\overline{{1}}\)

(\(\displaystyle{3}^{{2}}={9}\equiv{1}\in\text{mod}{8}\), similar approach for 5,7). So, there does not exists any element of order \(\displaystyle{4}\in{\mathbb{Z}_{{8}}^{\ast}}\), hence \(\displaystyle{\mathbb{Z}_{{8}}^{\ast}}\) is not cyclic.

\(\displaystyle{\mathbb{Z}_{{5}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{a}{e}{r}{e}{\gcd{{\left({a},{5}\right)}}}={1},{1}\le{a}\le{5}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{2}},\overline{{3}},\overline{{4}}\right\rbrace}.\)

Check that

\(\displaystyle{\left(\overline{{2}}\right)}^{{4}}={\left(\overline{{4}}\right)}^{{2}}={1},\)

\(\displaystyle{\left({4}^{{2}}={16}\equiv\in\text{mod}{5}\right)}\)

As here 4 is the element positive integer such that \(\displaystyle\overline{{2}}^{{4}}=\overline{{1}}\), therefore, order of \(\displaystyle\overline{{2}}\ne{\mathbb{Z}_{{5}}^{\ast}}={4}\). Hence, \(\displaystyle{\mathbb{Z}_{{5}}^{\ast}}\) is cyclic

2) Again \(\displaystyle{\mathbb{Z}_{{8}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{a}{e}{r}{e}{\gcd{{\left({a},{8}\right)}}}={1},{1}\le{a}\le{8}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{3}},\overline{{5}},\overline{{7}}\right\rbrace}.,\)

but \(\displaystyle\overline{{3}}^{{2}}=\overline{^}{2}=\overline{{7}}^{{2}}=\overline{{1}}\)

(\(\displaystyle{3}^{{2}}={9}\equiv{1}\in\text{mod}{8}\), similar approach for 5,7). So, there does not exists any element of order \(\displaystyle{4}\in{\mathbb{Z}_{{8}}^{\ast}}\), hence \(\displaystyle{\mathbb{Z}_{{8}}^{\ast}}\) is not cyclic.