How to determine one class of groups with order $p^{3}$? 
$G$ is a nonabelian group of order $p^{3}$,and $G=(Z_{p}\times Z_{p})\rtimes_{\theta} Z_{p}$, prove that: $$G\cong \begin{bmatrix}1&a&b\\0&1&0\\0&0&1\end{bmatrix}\rtimes\begin{bmatrix}1&0&0\\0&1&c\\0&0&1\end{bmatrix}$$ which can be written as internal semidirect product $G\cong (Z_{p}\times Z_{p})\rtimes Z_{p}$.

Attempt:
Firstly,for each element in $Z_{p}$, we can associate it with an element in $Aut(Z_{p}\times Z_{p})\cong GL(2,Z_{p})$. That is what $\theta$ all about. In our internal case the corresponding automorphism is just group of conjugate. In order to determine that two semidirect product have essentially the same structure,we have to dig into the automorphisms given above. But I don't know how to go further?
 A: Suppose the action in the semi-direct product is non-trivial.
As you rightly say we have $\theta: \mathbb{Z}_p\to \text{GL}(2,p)$; moreover everything is determined by $\theta(1)$ as $\theta(k)=\theta(1)^k$.   
Now as $p=0$ we must have that $\theta(1)^p=\theta(p)=\theta(0)=1$. By the non-triviality, $\theta(1)$ is of order $p$, and so its minimal polynomial divides $X^p-1= (X-1)^p$. As we are in $\text{GL}(2,p)$ the minimal polynomial is either $X-1$ (no, as $\theta(1)\not=1$), or $(X-1)^2$. Then $\theta(1)$ is by the Jordan Canonical Form Theorem similar to the matrix $\left[\begin{matrix}1 &1\\0&1\end{matrix}\right]$. 
Putting that another way, we can choose to write the $\mathbb{Z}_p\oplus\mathbb{Z}_p$ which is the normal subgroup of the semidirect product with respect to a basis such that chosen generator of the other $\mathbb{Z}_p$ acts like the matrix above. 
Using this basis, each element of $G$ has the form $((a,b),c)$. Map this to the matrix $\left[\begin{matrix}1 & a & b\\0 & 1 &c\\0 & 0 & 1\end{matrix}\right]$. It is now routine to check that this map is an isomorphism from $G$ (with its multiplication given by $\theta$) onto the group of lower triangular matrices (with its usual matrix multiplication).
A: Although I think something in your question doesn't seem correct, let me guess what is missing.
To start with, I think you should check the definition of what you call internal semidirect product. Now, up to isomorphism there are only two non-abelian groups of order $p^3$. The group of matrices with 1's along the diagonal over the field of $p$-elements $UT_3(\mathbb{F}_p)$, the so-called group of unipotent matrices or Heisenberg group, whilst the second one looks like $\mathbb{Z}_{p^2} \rtimes \mathbb{Z}_p$. The group you' re looking for is the first one, since you are talking about the internal semidirect product, you have to play around the definition of it and some elementary facts from group theory. For instance in your case, you can take advantage of the fact that for any $p$-group of cardinality $p^i$ where each element is of order $p$ there is a subgroup isomorphic with $C_p \times C_p$, proof of which can be found here Do groups of order $p^3$ have subgroups of order $p^2$?, moreover because the center of that group has order $p$, (a proof of this is again in the same answer but is a standard theorem that the center in that case is non-trivial), the intersection of these two subgroups has to be trivial (the first subgroup doesn't contain any element which commutes with the others of the group). But the latter implies your assertion if you look carefully the definition. I hope my answer helped you out.
Cheers!
