Source: Abstract Algebra, $3^{rd}$ edition by Dummit and Foote.
In Section 5.5, an example goes as follows:
Suppose $|G|=pq$ where $p, q$ are primes and $p<q$, $p|q-1$. Let $P\in Syl_p(G)$ and $Q\in Syl_q(G)$. Since $Aut(Q)$ is cyclic it contains a unique subgroup of order p, say $\langle\gamma\rangle$, and any homomorphism $\varphi: P\to Aut(Q)$ must map $y$ to a power of $\gamma$. There are therefore $p$ homomorphisms $\varphi_i: P\to Aut(Q)$ given by $\varphi_i(y)=\gamma^i$, $0\leq i\leq p-1$. Each $\mathbf{\varphi_i}$ gives rise to a non-abelian group, $\mathbf{G_i}$, of order $\mathbf{pq}$.
My Question: I understand that the group $G_i$ in the boldface texts refers to the semi-direct product $Q\rtimes_{\varphi_i}P$; I also know that there exists a non-abelian group of order $pq$. But how exactly does one see that $G_i$ is non-abelian?
Any hint would be greatly appreciated.
If $P$ and $Q$ are abelian and the operation in the group $G=Q\rtimes_\phi P$ is defined by the rule $$ (a,b)\cdot(a',b')=(a\phi(b)(a'),bb'), $$ where $a,a'\in Q$ and $b,b'\in P$, then the group $G$ is abelian if and only if $\phi(b)=1$ for all $b\in P$.
Compare $$ (a,e)\cdot(e,b)=(a\phi(e)(e),b)=(a,b), $$ $$ (e,b)\cdot(a,e)=(\phi(b)(a),b). $$ It follows that all your groups $G_i$ are non-abelian.