If $k$ is element of kernel of mapping $ f$, should it always be $f(k) = 0$?

Question

In an exercise there the is the following set:

$G = \left\{\begin{pmatrix}a&b\\ \:0&c\end{pmatrix}\mid a,b,c ∈ ℝ , ac ≠ 0 \right\}$

And there is the following mapping:

$Φ : G → ℝ \setminus\left\{0\right\}, \begin{pmatrix}a&b\\ \:0&c\end{pmatrix} → \det\begin{pmatrix}a&b\\ \:0&c\end{pmatrix}$

At one point of the exercise I had to find the kernel of Φ. I got stuck, because I though that $\ker Φ = \left\{0\right\}$, because the $0$ is not element of the $\operatorname{im}Φ$. And the determinant of any $g ∈ G$ cannot be $0$, because $ac ≠ 0$, anyway.

But to solve the problem, there was a kernel needed, so I checked the official solution. The following was written: $\ker Φ = \left\{k ∈ G \mid \det(k) = 1\right\}$

I don't understand. If $k ∈ \ker Φ$, should not be $Φ(k)=0$?

Answer 1

Your $\Phi$ is not a linear transformation, but a group homomorphsim from $G$ (which is a group under matrix multiplication -- you should prove this!) to the group of nonzero real numbers under multiplication.

The kernel of $\Phi$ then consists of those elements that map to the identity element of the group $\mathbb R\setminus\{0\}$, which is $1$.