And now, a stumper

Ok, this one's stumping me. Here goes:
Let G be a nontrivial group with no nontrivial proper subgroups.
Claim: |G| is prime.
Sidebar: Here's the problem. Well, the first one. It's easy to say that, if G is of prime order p, then the possible orders of its subgroups are either 1 or p. However, what's there to stop a group G of non-prime order n from not having any nontrivial proper subgroups? This relates back to what I was saying in my last post about how the converse of Lagrange's Theorem is not true. So, I really need to show that, if a group G is of non-prime order n, then it MUST have a nontrivial proper subgroup. Oh, wait. Let H be a subgroup of G, whose order is n and non-prime, and let a belong to G, where |a|=m and m does not equal n; by a corollary of Lagrange's Theorem, m|n. Then, 'a' is a nontrivial proper subgroup of G. So, G MUST have a nontrivial proper subgroup. However, if G is of prime order p, then all of its elements must be of order 1 or of order p, by the same corollary, so 'a'=G or 'a'=e, for all a in G. Great. Excellent. Half down. Now, the second problem. What if G is infinite? Obviously, I need to prove that G CANNOT BE infinite. Then, if it can't be infinite, it must be finite, and, as I've already showed, if it's finite, it's of prime order. So. How do I prove that it can't be infinite? Same way I showed that if its finite, its of prime order. That is to say, I need to show that any infinite group must have a nontrivial proper subgroup. How do I do this? Let's assume that G is an infinite group and that a belongs to G. Then, 'a' must be infinite as well. No help there, right? How about this; it's a little edgy, but I'll give it a go. Let G be an infinite group, and let a belong to G. Then, a possible subgroup of G is {e, a, a^-1}; this is a subgroup because it includes the identity and the inverse of each element, and it's closed. Note that, if G is of prime order, this subgroup doesn't fly: the subgroup has order 3, and, by Lagrange's Theorem, the order of the group is a multiple of the order of the subgroup, so the order of G cannot be prime, which contradicts our assumption. Ok. This feels a little sketch. But I'll go ahead and write it up formally now:

Claim: If G is a nontrivial group with no nontrivial proper subgroups, |G| is prime. \
Pf: Assume that G is a nontrivial group with no nontrivial proper subgroups. First, I need to show that G cannot be infinite. I will do this by contradiction. Assume that G is infinite, that a belongs to G, and |a}=2. Then, define a subgroup H, where H={e, a}. Then, H is a nontrivial proper subgroup of G, which contradicts our assumption that G has no nontrivial proper subgroups. So, G cannot be infinite.
Now, I need to show that |G| is prime. I will do this by contradiction again. Assume that |G| is non-prime. Let a be a non-identity element of G, let |G|=n, let |a|=m, and let m and n be distinct. Then, 'a' is a subgroup of G (I won't prove this; it's in my text as Thm 3.4, so I feel free to call upon it at will). However, because m and n are distinct, 'a' does not equal G (note that, if the orders of G and a are equal, 'a'=G). Then, 'a' is a nontrivial proper subgroup of G, contradicting our assumption that G has no nontrivial proper subgroups. Therefore, |G| must be prime. QED.
Ok, here's the problem I see with this proof: In a finite group G with non-prime order n, what if there are no elements that are of orders other than 1 and n? That is to say, what's stopping a group of order n having 1 identity and then n-1 elements of order n? That's a problem. I kinda just ran myself in a circle, maybe. The first half is solid, though. I'll keep thinking on it.