Proof showing that a group must be finite of prime order I'm trying to prove the following:
G is a group with order $\ge 2$ with no proper, non-trivial subgroups.  G must be finite of prime order.
My attempt:
Consider $g \neq e \in G$ (we can do this since order of $G$ is at least 2).  Since $G$ has no proper, non-trivial subgroups, $<g>$ can't be a proper subgroup of $G$.  Since it clearly can't be $e$, we must have $<g> = G$.
I'm not sure why it has to be finite though...
Help?
Thanks guys,
Mariogs
 A: So you have $\langle g\rangle=G$. If $g$ has infinite order, then $\langle g^2\rangle$ is proper in $G$ for instance, since $g\notin\langle g^2\rangle$. If it where, you'd have $g=g^{2k}$ for some $k$, or $g^{2k-1}=e$, contradiction. 
So $G$ is a finite cyclic group. Recall that a finite cyclic group has a unique subgroup of every order dividing $|G|$. This forces $|G|$ to be prime, otherwise you'd have a nontrivial, proper subgroup.
A: I suppose this question is from Herstein. If You have solved the previous one you know that in a group G if  intersection of all it's subgroup different from  is a subgroup different from  then all of it's elements have finite order. Here indeed this is the case.
A: |G|>=2 then G has a nonidentity element say a . Consider the sub group generated by a is . as G has no nontrivial sub group therefore G= .now G is cyclic . if |G| is infinite then G is isomorphic to Z .but Z has non trivial sub groups .therefore order of G is finite say n .now G to be cyclic ,G has a subgroup for each divisor of n . but  G has no nontrivial sub group .therefore divisor has to be 1 or n thus n is prime.
