Let $m$, $n$ be primes with $m < n$, $m \neq n \text{ (mod $1$)}$, and suppose $|G| = mn$. Prove that $G$ is cyclic.
The first thing I need to show is that there is only one Sylow $m$-subgroup and one Sylow $n$-subgroup. By the third Sylow theorem, we have: \begin{align*} n_m \equiv 1 \text{ (mod $m$)} \\ n_m \mid n \end{align*} The second condition implies that $n_m = n$ or $n_m = 1$. I should be able to rule out $n_m = n$, but I cannot figure out how. If $n_m = n$, then $n \equiv 1 \text{ (mod $m$)}$, so $n = 1 + mk$ for integer $k$. I doI qm wn't see how this produces a contradiction.
On the flip side: \begin{align*} n_n \equiv 1 \text{ (mod $m$)} \\ n_n \mid m. \end{align*} I am still not able to conclude $n_m = n_n = 1$.
Help would be appreciated.
It's obviously false, as $S_3$ is a non-cyclic group of order $6=2\cdot 3$.
You need another condition for this to be true, such as $m\nmid n-1$ for example.
In this case, you can prove there is a unique $n$-Sylow (you are almost there). If $H$ is the unique $n$-Sylow subgroup and $K$ is a $m$-Sylow subgroup, the next step is to prove that $G=H\rtimes K$. Now the corresponding morphism $K\to Aut(H)$ is trivial because $m$ is prime to $n-1$ (why ???). Then $G=H \odot K\simeq H\times K$, and CRT (for example) will lead to the desired conlusion.
Details are left to you.