 # 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

# Knowledge Graph # 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

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