Question

How do you prove that if a group `G` has no non-trivial subgroups, then it is finite of prime order?

Answer 1

Consider a group `G`.

It's known that `G` has no non-trivial subgroups.

For `G` to have a finite order, it must be a finite group.

Thus the only subgroups are, `G` itself and `{i}` where `i` is the identity element.

But, from Lagrange's theorem, order of any subgroup must divide the order of `G`.

If order of `G` be `o(G) = x` then of the subgroups of `G`, the orders are `x` and `1` only. (Because order of `{i}` is `1`)

Thus, order of `o(G) = x` has only two factors which are `1` and itself `x`.

Hence `x` is prime and `G` is a group of prime order and is finite.

Answer 2

The proposition is false, since the conditions hold for the group of one element. But if the group is non-trivial then it is isomorphic to `C_p` for some prime `p`.

Explanation:

The proposition is false in that the group of one element satisfies the condition of having no non-trivial subgroups, but has non-prime order.

If we strengthen the conditions to exclude the group with one element, then let `a` be any non-identity element.

The subgroup generated by `a` must be the whole group.

If there is no positive integer `n` for which `a^n = 1`, then the element `a^2` generates a proper subgroup, since `(a^2)^m = a` would imply that `a^(2m-1) = 1`. So `a^n = 1` for some positive integer `n`.

If `n` is composite then choose a factor `m` such that `1 < m < n`.

Then we find that `a^m` generates a subgroup of order `n/m`, contradicting the condition that the group have no non-trivial subgroups.

Hence `n` is prime and the group is isomorphic to `C_p` for some prime `p`.