Here is the question I was reading:
Let $ p $ be an odd prime. Let $ H = Z_p \times Z_p $ and $ K=Z_p $. We know that the automorphism group $ \text{Aut}(H) $ is isomorphic to a general linear group, $ \text{Aut}(H) \cong \text{GL}_2(F_p) $ and its size is $ |\text{GL}_2(F_p)| = (p^2-1)(p^2-p) $ thus $ p | \text{Aut}(H) $ and by Cauchy's theorem there is an element of $ \text{Aut}(H) $ of order p and hence a nontrivial group homomorphism $ \phi: K \to \text{Aut}(H) $ and so the associated semidirect product $ H \rtimes_{\phi} K $ is a non Abelian group of order $ p^3 $. More specifically, if $ H = \langle a \rangle \times \langle b \rangle $ and $ K = \langle x \rangle $ then $ x $ acts on $ a $ and $ b $ by:
$$ x \cdot a = ab $$ $$ x \cdot b = b $$
With respect to the $ F_p $ basis $ a,b $ of the 2-dimensional vector space $ H $ the action of $ x $ in addititve terms is the linear transform
$$ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} $$
It can be found in the following link:
Need help understanding example of semidirect product
My question is:
In the statement of the question above, why the associated semidirect product $H \rtimes_{\phi} K$ is a non Abelian group of order $P^3$, could anyone explain this to me, please?