Cyclic Groups & Subgroups : IIT 2018 Problem 1

Understand the problem

Which one of the following is TRUE? (A) \(\Bbb Z_n\) is cyclic if and only if n is prime
(B) Every proper subgroup of \(\Bbb Z_n\) is cyclic
(C) Every proper subgroup of \(S_4\) is cyclic
(D) If every proper subgroup of a group is cyclic, then the group is cyclic.

Start with hints

Hint 1:

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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Similar Problems

TIFR 2013 problem 5 | Non-Cyclic Subgroup of \(\mathbb{R}\)

Try this problem from TIFR 2013 problem 5 based on Non-Cyclic Subgroup of \(\mathbb{R}\).

Question: TIFR 2013 problem 5

True/False?

All non-trivial proper subgroups of \((\mathbb{R},+)\) are cyclic.

Hint: What subgroups comes to our mind immediately?

Discussion: \((\mathbb{Q},+)\) is a subgroup of \((\mathbb{R},+)\). Is \((\mathbb{Q},+)\) a cyclic group?

Suppose \((\mathbb{Q},+)\) is cyclic. Then there exists a generator say \(\frac{a}{b}\). Note that, we are only allowed to use addition (and subtraction) to create \(\mathbb{Q})\

Therefore, we can create $$ \frac{a}{b}+\frac{a}{b}+...+\frac{a}{b}=n\frac{a}{b}=\frac{na}{b} $$

Also, we can create  $$ (-\frac{a}{b})+(-\frac{a}{b})+...+(-\frac{a}{b})=n(-\frac{a}{b})=-\frac{na}{b} $$

Notice that we can increase the magnitude of the numerator, but not the magnitude of the denominator.

For example, we cannot create \(\frac{a}{2b}\) using \(\frac{a}{b}\) and the binary operation +.

Therefore, \((\mathbb{Q},+)\) is not cyclic.

Remark: There is one result which states that subgroups of \((\mathbb{R},+)\) are either cyclic or dense. Notice that although \((\mathbb{Q},+)\) is not cyclic it is dense.

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