Every infinite group $G$ has a subgroup $H$ that is non-trivial ($H \ne G, \lbrace e \rbrace$).
Proof: This will be a proof by contradiction. So we will assume every subgroup is trivial and bring the cyclic subgroups to the table.
For the non-identity $\forall x \in G$, $\langle x \rangle \ne \lbrace e \rbrace$. Hence $\langle x \rangle = G.$ Thus every non-identity element of $G$ must be a generator of $G$. Then it must be possible to write any element as the exponent of another. For $\forall y \in G$,
Since $x^2 \in G$ by closure, it is also a generator and we must be able to write $x$ as an exponent of $x^2$. Yet this is not possible unless our group is finite. So our assumption must be false and our conjecture must be true. $\square$
Is my proof watertight? And is it rigorous enough? Thanks for your reviews.