Let $K$ be a field and $$G :=\left\{\begin{pmatrix} a & b\\ 0 & d \end{pmatrix}\,: a,d \in K^*, b \in K\right\} \subset GL_2(K).$$ I want to show that $G$ is isomorphic to the semidirect product $K \rtimes (K^* \times K^*).$ I need to find two subgroups $G_1, G_2$ in $G$ having trivial intersection and one of them being normal, such that $G = G_1G_2.$
Finding the appropriate homomorphism is also not difficult. What I do get for $G_1, G_2$ is subgroups consisting of $2\times 2$ matrices. But I do not recognize such subgroups while looking at the given semidirect product. In fact, the field $K$ consists of scalors, i.e. $1 \times 1$ matrices and $K^* \times K^*$ of $1 \times 2$ matrices which I do not see how to consider them as subgroups of $G$.
How should I proceed?
Thanks for your suggestions.
Let the two subgroups be $$G_1=\left\{ \begin{pmatrix} 1&b\\&1\end{pmatrix}\middle| b\in K\right\}\ \text{ and }\ G_2=\left\{ \begin{pmatrix} a&\\&d\end{pmatrix}\middle| a,d\in K^*\right\}.$$ It is easy to see that....(1) $G_1$ is normal in $G$, (2) $G_1\cap G_2=\{I\}$, (3) $G_1G_2=G$. Moreover, the following maps $$\varphi_1\colon G_1\rightarrow K; \begin{pmatrix} 1&b\\&1\end{pmatrix}\mapsto b\ \text{ and }\ \varphi_2\colon G_2\rightarrow K^*\times K^*; \begin{pmatrix} a&\\&d\end{pmatrix}\mapsto (a,d)$$ can be seen to be isomorphisms of groups. This would clarify the problem.