# Prove the following.(1) Z ∗ 5 is a cyclic group. (2) Z ∗ 8 is not a cyclic group.

he298c 2021-02-27 Answered

Prove the following.
(1) $$Z \times 5$$ is a cyclic group.
(2) $$Z \times 8$$ is not a cyclic group.

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## Expert Answer

Isma Jimenez
Answered 2021-02-28 Author has 7605 answers

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,
$$\displaystyle{\mathbb{Z}_{{5}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{e}{r}{e} \ {\gcd{{\left({a},{5}\right)}}}={1},{1}\le{a}\le{5}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{2}},\overline{{3}},\overline{{4}}\right\rbrace}.$$
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 $$\displaystyle{\mathbb{Z}_{{8}}^{\ast}}={\left\lbrace\overline{{a}}:{w}{h}{e}{r}{e} \ {\gcd{{\left({a},{8}\right)}}}={1},{1}\le{a}\le{8}\right\rbrace}={\left\lbrace\overline{{1}},\overline{{3}},\overline{{5}},\overline{{7}}\right\rbrace}.,$$
but $$\bar{3}^2=\bar{5}^{2}=\bar{7}^2=\bar{1}$$
($$\displaystyle{3}^{{2}}={9}\equiv{1}\in\text{mod}{8}$$, 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.

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