I want to prove that if G has order pq, primes, with q>p and p DOES NOT divides q-1, then G is cyclic.
My attempt:
By Sylow's theorem,$n_p \equiv 1 (\textrm{mod}\ p)$ and $n_q \equiv 1 (\textrm{mod}\ q)$, and more, $n_p \equiv 0 (\textrm{mod}\ pq)$ and $n_q \equiv 0 (\textrm{mod}\ pq)$. From here we get that $n_p=1$ or q and $n_q=$1 or p, combining with what is given by the question, we see that $n_p=1=n_q$, therefore there are only one subgroup of this orders.
Let K be the one with order p and H with order q. Since both have a prime as order, they are cyclic. Then, only the neutral elemente lies in the intersection of the two, hence $G=K \times H$ because $|G|=pq$ and both K and H are normal, since they are unique of each order.
So,since p and q are coprimes, G is cyclic.
Is that correct? This seems just to simple and i am thinking i am making some silly mistake, but i cant see.
Thanks to everyone.
Best Answer
This is correct, and the normality of the Sylow subgroups is where the condition that $p$ not divide $q-1$ comes in. If $p$ does divide $q-1$, then we can have $q$ subgroups of order $p$ and a unique nonabelian group structure arises out of this.