= Solution
For either coefficient system $\mathbb Z$ or $\mathbb F_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the <determinant> preserves <determinant> one. The identity matrix has <determinant> one, and for a determinant-one matrix the <matrix inverse> is
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}
=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.
$$
Its entries again belong to the same coefficient system and its <determinant> is one. Together with associativity of <matrix multiplication>, these facts establish all the <group axioms>. Thus both sets are <special linear groups>.
Reduce each entry modulo five to define
$$
\pi:SL_2(\mathbb Z)\longrightarrow SL_2(\mathbb F_5),\qquad M\longmapsto\overline M.
$$
Reduction respects sums and products, so $\overline{MN}=\overline M\,\overline N$. It also respects the <determinant>, ensuring that the image has <determinant> one. Hence $\pi$ is a <group homomorphism>. Its <kernel of a group homomorphism> consists exactly of determinant-one integer matrices congruent to the identity modulo five:
$$
\boxed{\ker\pi=\left\{\begin{pmatrix}1+5a&5b\\5c&1+5d\end{pmatrix}\in SL_2(\mathbb Z):a,b,c,d\in\mathbb Z\right\}.}
$$
The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A <kernel of a group homomorphism> is a <normal subgroup>, since $\pi(gkg^{-1})=\pi(g)e\pi(g)^{-1}=e$ for $k\in\ker\pi$. This is the <principal congruence subgroup> of level five.
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