Problem: Let $N$ and $H$ be any group and let $\theta \colon H \rightarrow \text{Aut}(N)$ is a homomorphism. The the set $$G = \{(x,h) \mid x \in N, h \in H\},$$ is a group with operation defined as follow: for all $(x_1,h_1) \in G$ and $(x_2,h_2) \in G$, $$(x_1,h_1)(x_2,h_2) = (x_1 \theta_{h_1} (x_2), h_1 h_2).$$ Denote $N_0 = \{(x,1) \mid x \in N\}$ and $H_0 = \{(1,h) \mid h \in H\}$. Then $N_0 \cong N$, $H_0 \cong H$ and $G$ is an inner semidirect product of $N_0$ by $H_0$. Moreover $$(1,h)(x,1)(1,h)^{-1} = (\theta_h (x),1),$$ for all $x \in N$ and $h \in H$.
My proof: To prove that $G$ is a group, we have to check three conditions in the definition of a group.
Firstly, the operation is associative. $\forall (x_i,h_i) \in G, i = 1,2,3$. $$[(x_1,h_1)(x_2,h_2)](x_3,h_3) = [x_1 \theta_{h_1} (x_2), h_1 h_2](x_3,h_3) = (x_1 \theta_{h_1} (x_2) \theta_{h_1 h_2}(x_3),h_1 h_2 h_3)$$ $$(x_1,h_1)[(x_2,h_2)(x_3,h_3)] = (x_1,h_1)[x_2 \theta_{h_2} (x_3),h_2 h_3] = (x_1 \theta_{h_1}(x_2 \theta_{h_2}(x_3)),h_1 h_2 h_3) = (x_1 \theta_{h_1}(x_2) \theta_{h_1 h_2}(x_3),h_1 h_2 h_3) = [(x_1,h_1)(x_2,h_2)](x_3,h_3).$$
Secondly, the identity element. Let $e$ be an identity element of $G$, $e = (e_N,e_H)$ since for any $(x,h) \in G$ we have $$e (x,h) = (e_N,e_H)(x,h) = (e_N \theta_{e_H}(x),e_H h) = (e_N x, e_H h) = (x,h) = (x e_N, h e_H) = (x \theta_h(e_N),h e_H) = (x,h)(e_N,e_H).$$
Finally, $(x,h)^{-1} = ((\theta_{h^{-1}}(x))^{-1},h^{-1})$ since $$(x,h)((\theta_{h^{-1}}(x))^{-1},h^{-1}) = (x \theta_h (\theta_{h^{-1}}(x^{-1})),h h^{-1}) = (e_N,e_H)$$
To prove that $N_0,H_0$ are subgroups of $G$ and $N_0 \triangleleft G$, we do
$(x,1),(y,1) \in N_0, (x,1)(y,1) = (xy,1) \in N_0$, prove similarly for $H$, so we conclude that $N,H$ are subgroups of $G$.
$$(x,h)(x,1)(x^{-1},h^{-1}) = (x \theta_h(x),h)((\theta_h(x))^{-1},h^{-1}) = (x \theta_h(x) \theta_h ((\theta_h(x))^{-1}),1) = (x \theta_h (x) (x^{-1}),1) = (x,1),$$ so $N_0 \triangleleft G$.
$N_0 \cap H_0 = \{1\}$ since $1 = (1,1)$ is both in $N_0$ and $H_0$.
For each element $(x,h) \in G$ can be written uniquely in the form $(x,h) = (x,1)(1,h)$. So $G$ is an inner semidirect product of $N_0$ by $H_0$.
Please check my proof and do I have to prove that $(1,h)(x,1)(1,h)^{-1} = (\theta_h (x),1)$?