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

Why is it interesting to laminate a genus-2 surface?

 The term double torus is occasionally used to denote a genus 2 surface.

We are interested to understand the structure of a group G (is it abelian, or a product of abelian groups, or a free group of some kind etc.). In general, this is a (very) hard question. One strategy is to let G act on some well known (topological) space. Often by studying the effect of this group action on the space, it is possible to comment on the algebraic structure of the group.

Why this ‘indirect’ method should be regarded as ‘natural’? It is useful to think about elements of groups as ‘actors’. Their true color is revealed, only when they are unleashed in a stage (topological space).

However, we have to find the right space on which the group under scanner is to be let loose. A lot of effort goes into the construction (and investigation) of topological spaces which will be effective stages for group action.

Bass Serre theory produced a wonderful (topological) space that produces important information about groups acting on them. They are cleverly designed simplicial trees. We look at the stabilizers of vertices and edges of this tree. This process reveals a lot of information about the structure of the group.

Simplicial trees are different from (non-simplicial) R-trees (real trees). The key distinction is: R-trees have non-discrete branching points. There is an interesting construction in this context. Let G act isometrically (preserving distances), on a sequence of negatively curved spaces. Then we have a natural isometric action of G on an R-tree in the Gromov Hausdorff limit. 

This makes R-trees the final destination space of the isometric Group action. It makes sense to study the R-trees. After all, they are stage of action of the Group that is our ultimate object of interest. A process of resolution leads us from R-trees to (measured) laminated 2-complexes.

(Ref: Bestvina; 1999)

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