Cyclic and non proper subgroups

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1. What is/are the condition for a group with no proper subgroup to be cyclic?


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3. this is just a general qustion I am asking in oder to prove something?
 
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Let a be non-identity member of G. If for some n, an= e, then {a, a2, ..., an-1, an= e} is a subgroup of G.

Now if G has no proper subgroups what is the smallest such n for any a? What does that tell you about G?
 
I don't quite understand, but I'm guessing the smallest such n would be 1... can you give me an example of a Group with non proper subgroups
 
I think I figurd it out ...the smallest of such n would be distinct , the group itself would bconsist of elements of infinite order... I still need an example of a Cyclic group with non proper subgroups
 
Do you really understand what you are asking? Any group of prime order has no proper subgroup.
 
Not quite...Ok this is my problem:
If G has noroper subgroups, prove that G is cyclic.

Proof:
If G has no proper subgroup then |G|= p. For any nonidentity element a belonging to G, <a> is a subgroup of order greater than 1. By Langrange's Theorem, since |a|divides |G| |a| = p therefore, <a> = G and G is a cyclic group of order p.

Does this proof make sense? In your first question I sain that the smallest scuh n is 1 hence the reason why I said that |a| must be greater than one...I'm notre if my proof is correct but does it make sense?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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