Question

Prove that if G≠{e} and G has no proper non-trivial subgroup, then G is finite and o(G) is a prime number?

Answer 1

See explanation...

Explanation:

Since `G != { e }` there is some non-identity element `a in G`.

If `a` is of infinite order then `< a^2 >` is a proper non-trivial subgroup of `G`, since it does not contain `a`.

If `a` is of finite order, then `a^n = e` for some positive integer `n`, which without loss of generality we can take to be the smallest possible value.

If `n = pq` is composite, with `p, q > 1` then `< a^p >` is a proper cyclic subgroup of order `q`.

Hence we can deduce that `n` is prime.