Cosets and Lagrange's Thm?

Question

Suppose that G is a group with more than one element and G has no proper, nontrival subgroups. Prove |G| is prime. (Do not assume at the outset that G is finite).

Answer 1

Since G has more than one element, let a be an element of G not equal to the identity. Then consider , the subgroup of G generated by a. This is a nontrivial subgroup of G, and since G has no proper nontrivial subgroups, it follows that =G, so G is cyclic. Now, suppose it is the case that j≠k ⇒ a^j≠a^k. Then since 1 is odd, there is no n∈ℤ such that 2n=1, so ∄n s.t. a^(2n) = a. This means a∉, so is a proper nontrivial subgroup of G - a contradiction. Thus ∃j, k s.t. j≠k but a^j=a^k. Then a^(j-k) = a^(k-j) = e, where one of j-k and k-j are positive, so a has finite order, and it is easy to show that || = o(a). Since G=, it follows that |G| is finite. Further, we know that |G|≥2, so |G| is either prime or composite. Suppose |G| is composite, and let m be a factor of |G| other than 1 or |G|. For nm ≡ 1 mod |G| to have an integer solution in n, m would have to be coprime to G, and it is not, so this congruence has no integer solutions. But a^(nm) = a ⇔ a^(nm-1) = e ⇔ o(a) | (nm-1) ⇔ |G| | (nm-1) ⇔ nm-1 ≡ 0 mod |G| ⇔ nm ≡ 1 mod |G|, so ∄n such that a^(nm) = a, thus a∉. But, since m < o(a), a^m≠e, so is a nontrivial proper subgroup of G -- a contradiction. Therefore, |G| is not composite, so it must be prime. Q.E.D.