On a class of groups of order $p^2q$ Let $|G|=p^2q$ with following conditions:


*

*Sylow-$p$ subgroup is normal and is $\langle x,y\rangle \cong \mathbb{Z}_p\times \mathbb{Z}_p$.

*Sylow-$q$ subgroup is $\langle z\rangle$, and is not normal.

*No subgroup of order $p$ is normal in $G$.
Question: How many isomorphism types of such groups are? 
For $p=2$ and $q=3$, we know that there is such a group isomorphic to $A_4$.

While giving classification of groups of order $p^2q$ in a different way, I stuck at this case.
 A: By Sylow theorem, there is only $1$ Sylow $p$-subgroup which is $\Bbb{Z}_p\times \Bbb{Z}_p$. 
And by Sylow theorem again, the number of Sylow $q$-subgroup $n_q|1,\:p,\:p^2$, and $n_q\equiv1\mod q$. 


*

*If $n_q=1$, then there is only one Sylow $q$-subgroup which is normal. So this is impossible.

*If $n_q=p$, then there are $p$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup. So there are at most $p(q-1)+p^2$ elements. But since
$$p^2q-(p(q-1)+p^2)=p(q-1)(p-1)>0$$
total elements are less than $p^2q$, which is impossible. 
So only possibility is $n_q=p^2$, and there are $p^2$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup of $\Bbb{Z}_p\times \Bbb{Z}_p$ that is normal. Since there are more than one Sylow $p$-subgroup, $G$ is non-abelian. We prove that it is the only isomorphism type of $G$.
Suppose $\:H=\Bbb{Z}_p\times \Bbb{Z}_p=\{h:h^p=1,\:h\in H\}$ and $K=\Bbb{Z}_q=\{k:k^q=1\}$. Clearly $K$ is cyclic. Since $H\vartriangleleft G, \:kHk^{-1}=H, \:\forall k\in K$. Any conjugate mapping $\{khk^{−1}=h_1,h,h_1∈H, k\in K\}$ can be made isomorphic by inner automorphism as $\{\phi:\:\phi(x)=g^{-1}xg,\:x\in G,\:g\in G, \: g\ne 1, \:g\notin Z(G)\}$ for the same center.
Since isomorphism keeps center of group invariant, the number of isomorphism types depends on the number of center of $G$. Since $Z(G)\vartriangleleft G$ and $|Z(G)|\mid|G|$, $|Z(G)|=1,q,p,p^2$. By given condition, $|Z(G)|\ne p$. If $|Z(G)|=p^2$, then $|G/Z(G)|=q$. So $G/Z(G)$ is cyclic and $G$ is abelian, which is contradiction. $|Z(G)|=q$ is also impossible for none of Sylow $q$-subgroups is normal. So $|Z(G)|=1$. 
So there is only one isomorphism type for $G$ with $p^2$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup of $\Bbb{Z}_p\times \Bbb{Z}_p$.
