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

1) We know that Zn∗={a―:where gcd(a,n)=1,1≤a≤n} form a froup under multiplication with inverse bar1. Now, Z5∗={a―:where gcd(a,5)=1,1≤a≤5}={1―,2―,3―,4―}. Check that (2―)4=(4―)2=1, (42=16≡∈mod5) As here 4 is the element positive integer such that 2―4=1―, therefore, order of 2―≠Z5∗=4. Hence, Z5∗ is cyclic 2) Again Z8∗={a―:where gcd(a,8)=1,1≤a≤8}={1―,3―,5―,7―}., but 3¯2=5¯2=7¯2=1¯ (32=9≡1∈mod8, similar approach for 5,7). So, there does not exists any element of order 4∈Z8∗, hence Z8∗ is not cyclic.

Answered: "Prove the following. (1) Z×5 is a cyclic..."

College
Algebra
Abstract algebra

he298c

Answered question

2021-02-27

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

Answer & Explanation

Isma Jimenez

Skilled2021-02-28Added 84 answers

1) We know that $Z_{n}^{*} = {\overset{―}{a} : w h e r e gcd (a, n) = 1, 1 \leq a \leq n}$ form a froup under multiplication with inverse bar1. Now,
$Z_{5}^{*} = {\overset{―}{a} : w h e r e gcd (a, 5) = 1, 1 \leq a \leq 5} = {\overset{―}{1}, \overset{―}{2}, \overset{―}{3}, \overset{―}{4}} .$
Check that
${(\overset{―}{2})}^{4} = {(\overset{―}{4})}^{2} = 1,$
$(4^{2} = 16 \equiv\in mod 5)$
As here 4 is the element positive integer such that ${\overset{―}{2}}^{4} = \overset{―}{1}$ , therefore, order of $\overset{―}{2} \neq Z_{5}^{*} = 4$ . Hence, $Z_{5}^{*}$ is cyclic
2) Again $Z_{8}^{*} = {\overset{―}{a} : w h e r e gcd (a, 8) = 1, 1 \leq a \leq 8} = {\overset{―}{1}, \overset{―}{3}, \overset{―}{5}, \overset{―}{7}} .,$
but ${\bar{3}}^{2} = {\bar{5}}^{2} = {\bar{7}}^{2} = \bar{1}$
( $3^{2} = 9 \equiv 1 \in mod 8$ , similar approach for 5,7). So, there does not exists any element of order $4 \in Z_{8}^{*}$ , hence $Z_{8}^{*}$ is not cyclic.

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