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# Cyclic Groups & Subgroups : IIT 2018 Problem 1

This is an application abstract algebra question that appeared in IIT JAM 2018. The concept required is the cyclic groups , subgroups and proper subgroups.

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# What We Are Learning?

Groups are the main concept in abstract algebra here we will see about some application of subgroups and cyclic groups

# 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.
IIT Jam 2018
##### Topic
Groups , Cyclic Group & Proper Subgroup
EASY
##### Suggested Book
ABSTRACT ALGEBRA BY DUMMIT AND FOOTE

Do you really need a hint? Try it first!

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)
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)?

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.

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

# Some interesting Fact

Do you know that a cyclic group $\Bbb Z_n$ can be seen inside a circle $<e^{\frac{2\pi i}{n}}>$? Below is one picture of $\Bbb Z_8$ in the circle…

# Connected Program at Cheenta

#### College Mathematics Program

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.

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