Question

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

Abstract algebra
ANSWERED
asked 2021-02-27

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

Answers (1)

2021-02-28

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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