Let $G=GL_n(\mathbb{R})$, the group of invertible $n\times n$ real matrices,
Let $N$ be the subset of $G$ given by $\{A\in G:\det(A)>0\}$
Let $H=\{1,-1\}$ considered as a group under multiplication
Question: Find a homomorphism $\varphi:G\rightarrow H$ with $\ker(\varphi)=N$
and deduce that $N\unlhd G$ and $G/N\cong H$.
Attempt: The map has to map any matrix with a positive determinant to 1 and any matrix with a zero or negative determinant to $-1$, since $\ker(\varphi)=N$ this can be given by:
$$\varphi(g) = \left\{ \begin{array}{ll} 1 & \mbox{if $\det g > 0$};\\ -1 & \mbox{if $\det g \leq 0$}.\end{array} \right.$$
The homomorphism property is inherent from $\det(AB)=\det(A)\det(B)$ for $A$ and $B$ being $n\times n$ real matrices
To deduce that $N=\ker(\varphi)\unlhd G$,
Let $x\in \ker(\varphi)$ and $g\in G$, need to show $gxg^{-1}\in\ker(\varphi)$
$\varphi(x)=\{1\}$
We know $\{1\}\unlhd H$, therefore, $h\cdot 1\cdot h^{-1}\in\{1\}$ for $h\in H$
Therefore letting $\varphi(g)=h$, we get $\varphi(g)\varphi(x)\varphi(g)^{-1}\in\{1\}$
Since $\varphi$ is a homomorphism, we have $\varphi(gxg^{-1})\in\{1\}$ and $gxg^{-1}\in\varphi^{-1}(\{1\})=\ker(\varphi)$
Therefore $N=\ker(\varphi)\unlhd G$
The $G/N\cong H$ property is clear from the homomorphism theorem which states:
$G/\ker(\varphi)\cong\text{im}(\varphi)$, since $N=\ker(\varphi)$ and im$(\varphi)=H$. Therefore this holds.
Problems: Is this map correct, how can I improve this answer and if my answer is correct does there exist another map which would fit this question?