Let $K$ be a field, $H=\left\{\begin{bmatrix}a&b\\0&d\end{bmatrix}:a,b,d\in K, ad\ne0\right\}$, $L=\left\{\begin{bmatrix}1&b\\0&1\end{bmatrix}:b\in K\right\}.$
I'm asked to prove that $H/L$ is isomorphic to $K^*\times K^*$, where "$K^*=(K_{\ne0},\cdot)$". Now, this means I should find a bijective $f:H/L\to K^*\times K^*$ such that $f(AB)=f(A)\cdot f(B)$ - my first doubt is that the operator "$\cdot$" is associated with the sole $K_{\ne0}$ rather than $K_{\ne0}\times K_{\ne0}$. Now, given that $H/L$ is isomorphic to $U=\left\{\begin{bmatrix}a&0\\[]0&d\end{bmatrix}:a,d\in K, a\ne d, ad\ne0\right\}$, am I correct in saying that $f:\begin{bmatrix}a&0\\[]0&d\end{bmatrix}\mapsto(a,d)$ does the job because $$f\left(\begin{bmatrix}a_1&0\\[]0&d_1\end{bmatrix} \begin{bmatrix}a_2&0\\[]0&d_2\end{bmatrix}\right)=f\left(\begin{bmatrix}a_1a_2&0\\[]0&d_1d_2\end{bmatrix}\right)=(a_1a_2,d_1d_2)=(a_1,d_1)\cdot(a_2,d_2)$$? I also have another concern: it seems to me that the $a\ne d$ bit in $U$ makes the cardinality of $U$ shift from that of $K_{\ne0}\times K_{\ne0}$ (at least if $K$ is finite), so I guess it shouldn't be there? But why?