Let $G=\langle a,b |a^q=1,b^{p^2}=1, bab^{−1}=a^i, ord_q(i)=p^2\rangle$, here $p^2 \mid q-1$ and $p< q$ are primes and I am assuming that $|G|>12$. Thus $G$ has a normal sylow $q$-subgroup.
How can I show $G$ has no element of order $pq$?
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Consider the general element $a^mb^n$, with $0\le m<q$ and $0<n<p^2$. We can assume $n\neq0$, since if $n=0$, this "general element" has order $q$.
Note that $$ (a^mb^n)^2 = a^m(b^na^mb^{-n})b^{2n}$$ and that thing in the middle is $a^{mk}$, with $k=i^n$. So our product is $a^{m(k+1)}b^{2n}$.
Doing this $p$ times gives $$ (a^mb^n)^p = a^{rm}b^{pn}$$ where $r=1+i^n+i^{2n}+\ldots i^{(p-1)n}$.
For $a^mb^n$ to have order $pq$, the element above must be in the unique Sylow $q$-subgroup of $G$; that is, we need $b^{pn}=1$, so that $p|n$.
But $r=\dfrac{i^{pn}-1}{i^n-1}$, and if $p|n$, the we have $r\equiv0\pmod{q}$. In other words, $(a^mb^n)^p=1$.
Steve D
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