Question

Prove that if G is not equal to {e} and G has no proper non-trivial subgroup,then G is finite and o(G) is a prime number.?

Answer 1

If `G` is infinite it has infinitely many subgroups, so it can't be infinite.

If `o(G)` is not prime, there exists `g \in G`, whose order divides `o(G)` that generates a non-trivial subgroup.

Explanation:

`G` must be finite, because there's a theorem stating that

Finite number of subgroups `\implies` finite group

So, using `(A\implies B) \implies (\not B \implies \not A)`

we deduce that an infinite group has an infinite number of subsets, and so `G` must be finite, otherwise it would have subgroups and we are assuming it doesn't.

Now, let's prove that the order of `G` must be prime: the order of every element `g \in G` divides the order of the group. So, if `o(G)` is not prime, it has proper divisors, say `k|o(G)`, and `k\ne 1` and `k \ne o(G)`.

So, there exists an element `g \in G` with `o(g)=k`, which would define a subgroup generated by powers of `g`.

Instead, if `o(G)` is prime, the order of every `g \in G` can either be `1` or `o(G)`. If the order is `1` the element is `e`, and the subgroup it generates is the trivial one.

If the order is `o(G)`, the subgroup generated is the whole group `G`. So, the only subgroups are the trivial ones.