We will solve this question by the method of elimination. Observe that if n is prime then \(\mathbb{Z}_n\) is obviously cyclic as any of the subgroup <a> has order either 1 or n by Lagrange's theorem. Now if the order is 1 then a=id. So choose a(\(\neq\)e) \(\in \mathbb{Z}_n\) then |<a>|=n and <a> \(\subseteq\) \(\mathbb{Z}_n\) \(\Rightarrow\) <a>= \(\mathbb{Z}_n\). The problem will occur with the converse see \(\mathbb{Z}_6\) is cyclic but 6 is not prime. In general \(\mathbb{Z}_n\) = <\(\overline{1}\)> is always cyclic no matter what n is!! so option (A) is false. Can you rule out option (C)
Hint 2:Consider option (C) every proper subgroup of \(S_4\) is cyclic. Consider { e , (12)(34) , (13)(24) , (14)(23) } = G Observe that this is a subgroup and |G|=4. Moreover o(g)=2 \(\forall\) g(\(\neq\)e) \(\in\) G So G is not cyclic. Hence option (C) is not correct. Can you rule out option (D)?
Hint 3:Consider \(\mathbb{Z}_2\)*\(\mathbb{Z}_2\) which is also known as Klein's 4 group then it is not cyclic but all of it's proper subgroups are {0}*\(\mathbb{Z}_2\) , \(\mathbb{Z}_2\)*{0} and {0}*{0} which are cyclic. Hence we can rule out option (D) as well.
Hint 4:So option (B) is correct. Now let prove that H \(\leq\) \(\mathbb{Z}_n\) = {\(\overline{0}\),\(\overline{1}\),.....,\(\overline{n-1}\)}. By well ordering principle H has a minimal non zero element 'm'. Claim: H=<m> clearly <m> \(\subset\) H. For any r \(\in\) H by Euclid's algorithm we have r=km+d where 0 \(\leq\) d < m which \(\Rightarrow\) d=r-km \(\in\) H If d \(\neq\) 0 then d<m which is a contradiction So, d=0 \(\Rightarrow\) r=km \(\Rightarrow\) H=<m> and we are done
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