• Feb 3rd 2010, 04:13 PM
ChrisBickle
Okay here is the question and ill give the answer i put i dont know if its sufficient or if there was another way.

Given G is a finite group whose only subgroups are {e} and G itself

A)Show G is cyclic

2 choices for G either G = {e} and G is cyclic. If not then there are more elements in G and since the only subgroups are the ones mentioned these elements must generate the entire group or else there would be another subgroup so there is a generator and G is cyclic.

B)Show G is of prime order

G is cyclic so G is isomorphic to Zn then we know G has subgroups generated by elements that divide order of G. There are only those 2 so order G is only divisible by 1 and itself so n is prime
• Feb 3rd 2010, 05:37 PM
Drexel28
Quote:

Originally Posted by ChrisBickle
Okay here is the question and ill give the answer i put i dont know if its sufficient or if there was another way.

Given G is a finite group whose only subgroups are {e} and G itself

A)Show G is cyclic

If $\displaystyle G=\{e\}$ we are done. Otherwise, let $\displaystyle g\in G$ then clearly $\displaystyle \left\langle g\right\rangle\leqslant G$ and since it's non-trivial it must be improper. Thus, $\displaystyle \left\langle g\right\rangle =G$

Quote:

B)Show G is of prime order

G is cyclic so G is isomorphic to Zn then we know G has subgroups generated by elements that divide order of G. There are only those 2 so order G is only divisible by 1 and itself so n is prime
That's good!