Let $p,q$ be primes with $p<q$ and $p \mid q-1$ . Show that there is a non-abelian group of order $pq$
By Sylow's theorems, we have $n_p=1 $ or $q$, and $n_q=1$.
However, if $n_p=1$, then both Sylow-p subgroup $P$ and Sylow-q subgroup $Q$ are unique and normal. By some work, we can see that $G$ is abelian.
So in order to show that there exists a non-abelian group of order $pq$, we must have $n_p=q$.
Thus, there are $q$ number of Sylow-p subgroups. Moreover, since $\gcd(q,p)=1$, for any Sylow-p subgroup $P$, we must have $P \cap Q = \{ e\}$.
From here, I'm a bit stuck on showing that $G$ is non-abelian.
I've seen answers that require the use of semi-direct product. But I wonder if there is a way without using it.
Thanks!
Best Answer
$\mathbb Z_q$ is a finite field, so its multiplicative group $\mathbb Z_q^{\times}$ is a cyclic group of order $q-1$. Since $p\mid q-1=|\mathbb Z_q^{\times}|$, there is a unique subgroup $P\leq \mathbb Z_q^{\times}$ satisfying $|P|=p$. Now consider the group of linear polynomial functions $G=\{ax+b\in \mathbb Z_q[x]\;|\;a\in P, b\in \mathbb Z_q\}$ under the operation of composition. By counting the number of choices for $a$ and $b$, derive that $|G|=pq$. Also, the linear function $t(x)=x+1\in G$ does not commute with $s(x)=ax\in G$ provided $a\neq 1$. There must exist such an $a\in P$, since $|P|=p>1$.