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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…

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