Groups and Subgroups!

1. Jul 10, 2008

vedicguru

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

we could say that, let a E G - {e} then we construct <a>...

2. Jul 10, 2008

quasar987

Sounds good. Why are you having doubts?

3. Jul 10, 2008

vedicguru

how do I construct <a> ??

4. Jul 10, 2008

HallsofIvy

Staff Emeritus
Well, you were the one who used that notation- what is the definition of "<a>"?

5. Jul 10, 2008

quasar987

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.

6. Jul 10, 2008

vedicguru

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!

7. Jul 10, 2008

quasar987

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".

8. Jul 10, 2008

vedicguru

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!