Proving G is Cyclic if No Subgroups Other than G and {e} Exist

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Homework Help Overview

The discussion revolves around proving that a group G is cyclic if it has no subgroups other than G and the trivial subgroup {e}. Participants explore the implications of this condition on the structure of the group.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the construction of the subgroup generated by an element a in G and its implications for the cyclic nature of G. Questions arise regarding the notation and definitions used in the context of group theory.

Discussion Status

The discussion is active, with participants clarifying definitions and exploring the reasoning behind the construction of the subgroup. Some guidance has been offered regarding the necessity of stating that the generated subgroup is not equal to {e} and its relationship to G.

Contextual Notes

There is an emphasis on ensuring that the definitions and properties of subgroups are correctly applied, as well as a focus on the implications of the group's structure based on the given conditions.

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Homework Statement



Prove that if a group G has no subgroup other than G and {e}, then G is cyclic...

Homework Equations





The Attempt at a Solution



we could say that, let a E G - {e} then we construct <a>...
 
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Sounds good. Why are you having doubts?
 
how do I construct <a> ??
 
Well, you were the one who used that notation- what is the definition of "<a>"?
 
Huh? Well by definition, <a>={..., a^-1,e,a,a²,a³,...}. Since <a> is a subgroup by construction and since it is not equal to {e}, it is equal to G.
 
oh yeah sure! I meant how would say, after constructing it! I forgot it is equal to {e} it is equal to G...this is what I get!
We could say that, let a E G - {e} then we construct <a> =
a, a^2, a^3... a^n=1 <- cyclic subgroup generated by element a.
Since <a> is subgroup, it must coincide with G (because G is the only subgroup), but then a generates G: <a> = G, which is definition of G being cyclic...
are we done after saying this statement or should we have to mention something else?
thanks for offering help!
 
Well what you wrote for the definition of <a> is incorrect, and don't forget to say "since <a> is not {e}, it is G".
 
ohh yeah I am sorry about that...I was in hurry or something to write it down! but yeah I believe everything else is fine and I would not forget to mention that <a> is not equal to {e}..it is G!
Thanks!
 

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