Show that G is cyclic

Question

Let G be a finite group with no subgroups apart from {1G} and G.
(a) Show that G is cyclic.
(b) Show that the number of elements in G is either 1 or a prime number.

Any ideas how can I solve the first question?

I was thinking to use Lagrange's theorem stating that the order of G must divide the order of {1G} and G. As G has no other subgroups, the order of G isn't divided by any other number. As the only numbers that have only 2 divisors (1 and itself) are prime numbers, therefore the order of G is a prime number. And we know that if the order of a group is a prime number then the group is cyclic.

Moreover, regarding the second one, is it enough to say that using (a) we know that a group is cyclic if its order is a prime number, therefore the number of elements in G is a prime number.

Answer 1

Suppose $G \neq \{e\}$. Let $g\neq e \in G$, and consider $\langle g \rangle$. Since $\langle g \rangle$ is a subgroup of $G$ and is nontrivial, it must therefore be all of $G$, hence $G$ is cyclic.

Your implication is not quite right. If $|G|$ is prime then $G$ is certainly cyclic, but the converse need not be true in general. You need to use specific information given. Instead, suppose $|G|$ were composite, and use Cauchy's theorem to find a subgroup that's nontrivial and not $G$.

If you don't have Cauchy's theorem at your disposal, or feel it is a bit heavy handed for this problem, then supposing $|G| = ab$ where $a, b > 1$, then if $G = \langle g \rangle$, then $$|g^{a}| = \gcd(ab, \frac{ab}{a}) = \gcd(ab, b) = b$$ so $\langle g^{a} \rangle$ has order...

Answer 2

If $G$ is trivial, then we are done. If $\exists x\ne1$. Then $\langle x \rangle$ is a nontrivial subgroup, and by the given condition, $\langle x \rangle=G$. So $G$ is cyclic.