Fake Proofs in Mathematics

By Cauchy's theorem, there exists an element $a\in G$ of order $p$ and an element $b\in G$ of order $q$.

Since $\gcd(p,q)=1$, the element $ab$ has order $pq$. Indeed, if $(ab)^m=e$, then both $p$ and $q$ must divide $m$, so $pq\mid m$. Therefore $|ab|=pq$.

Thus $G$ contains an element whose order is equal to $|G|$. Consequently, $G=\langle ab\rangle$, so $G$ is cyclic.