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
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
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
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
