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A subgroup $K$ is a normal subgroup of $G$ when $gK{g}^{-1}=K$ for every $g\in G$.

For a group homomorphism $\varphi :G\to H$, its kernel contains the identity and, for $x,y$ in the kernel, $\varphi (x{y}^{-1})=\varphi (x)\varphi (y{)}^{-1}=e$. Hence the kernel of a group homomorphism is a subgroup. Also $\varphi (gk{g}^{-1})=\varphi (g)e\varphi (g{)}^{-1}=e$ for $k$ in the kernel; applying this to $g$ and $g^{-1}$ gives equality under conjugation. Thus the kernel is always normal.

The image contains the identity and is closed under products and inverses, so the image of a group homomorphism is always a subgroup of the target. It need not be normal: include $C_2=\{e,(12)\}$ into $S_3$. Conjugation by $(123)$ takes $(12)$ to $(23)$, outside that image. Thus the image is not always normal.

For matrices over either $\mathbb{Z}$ or ${\mathbb{F}}_2={\mathbb{Z}}_2$, multiplication stays in the same coefficient ring, and determinants multiply, so the determinant-one matrices are closed. Associativity is inherited from matrix multiplication, the identity has determinant one, and the inverse isIts entries are still in the same ring and its determinant is one. This proves that both sets are special linear groups under multiplication. Over ${\mathbb{F}}_2$, minus and plus agree, and the same formula applies.

Reduction of integers modulo two preserves addition and multiplication. Therefore reducing entries preserves the determinant condition and satisfies $\varphi (AB)=\varphi (A)\varphi (B)$, proving the stated map is a group homomorphism.

Its target acts on the three nonzero vectors of ${\mathbb{F}}_2^2$. The action is faithful: a matrix fixing all three vectors fixes the two standard basis vectors and so is the identity. Thus its image, and in particular the image of $\varphi$, is isomorphic to a permutation subgroup of $S_3$. In fact SL2 over F2 as a permutation group identifies the entire image. There are $3$ choices for the first nonzero column and $2$ for a second independent column, hence six invertible matrices. Every invertible matrix has determinant one over ${\mathbb{F}}_2$, so the faithful action is onto $S_3$.

To show the integer reduction map reaches all of them, takeOn $\{e_1,e_2,e_1+e_2\}$, the reduction of $S$ exchanges $e_1,e_2$ and fixes their sum, while the reduction of $T$ fixes $e_1$ and exchanges the other two. These two transpositions generate $S_3$. Hence