= Solution
A <subgroup> $K$ is a <normal subgroup> of $G$ when $gKg^{-1}=K$ for every $g\in G$.
For a <group homomorphism> $\phi:G\to H$, its kernel contains the identity and, for $x,y$ in the kernel, $\phi(xy^{-1})=\phi(x)\phi(y)^{-1}=e$. Hence the <kernel of a group homomorphism> is a <subgroup>. Also $\phi(gkg^{-1})=\phi(g)e\phi(g)^{-1}=e$ for $k$ in the kernel; applying this to $g$ and $g^{-1}$ gives equality under conjugation. Thus \b[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 \b[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 is
$$
\boxed{\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}
=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\quad(ad-bc=1).}
$$
Its 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 $\phi(AB)=\phi(A)\phi(B)$, 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 $\phi$, 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, take
$$
S=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
T=\begin{pmatrix}1&1\\0&1\end{pmatrix}\quad\text{in }\mathrm{SL}_2(\mathbb Z).
$$
On $\{e_1,e_2,e_1+e_2\}$, 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
$$
\boxed{\operatorname{im}\phi=\mathrm{SL}_2(\mathbb F_2)\cong S_3.}
$$
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