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Let $G$ be a minimal non-abelian group with cyclic Sylow $p$-subgroup $P$ and normal Sylow $q$-subgroup $Q$ , see [ Huppert, Endlich Gruppen I, Aufgaben III, 5.14].
My quesion is, if there is another minimal non-abelian group $G_1$ which has the same order with $G$, then can we get $G\cong G_1$ ?
Thank you.
The answer is no, but there are at most two isomorphism types of minimal nonabelian groups of a given order that is divisible by $2$ distinct primes. (There are other examples that are $p$-groups such as $D_8$ and $Q_8$, but we are not talking about those here.)
Suppose $G$ is minimal nonabelian with $|G|=p^mq^n$, where $G$ has a cyclic Sylow $p$-subgroup $P$ and a normal Sylow $q$-subgroup $Q$. Then from the result that you quote, $Q$ is elementary abelian, and $P = \langle g \rangle$ is cyclic, with $g^p$ centralizing $Q$. Also $g$ acts irreducibly on $Q$, and so $p|(q^n-1)$ but $p$ does not divide $q^k-1$ for any $k<n$. So the Sylow $p$-subgroups of ${\rm GL}(n,q)$ are cyclic, and any two subgroups of ${\rm GL}(n,q)$ of order $p$ are conjugate. This implies that the isomorphism type of the semidirect product $Q \rtimes P$ is uniquely determined.
However, for some such orders we could reverse the roles of $p$ and $q$, giving two minimal nonabelian groups of that order. This happens for example with $|G| = 2^23$ or $|G|=3^45^2$.
