= Solution
The <first isomorphism theorem> says that for a homomorphism $\phi:G\to H$,
$$
G/\ker\phi\cong\operatorname{im}\phi,
\qquad g\ker\phi\mapsto\phi(g).
$$
The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.
Here $GL_n(\mathbb R)$ consists of invertible real $n\times n$ <matrices> and $SL_n(\mathbb R)$ consists of those with <determinant> one. The latter is the kernel of the surjective <determinant> map to $\mathbb R^*$, so it is normal and
$$
GL_n(\mathbb R)/SL_n(\mathbb R)\cong\mathbb R^*.
$$
The given <integral> <matrices> form a <group> because products and inverses remain <integral>. The inverse formula shows that an <integral> inverse exists exactly when the <determinant> divides every cofactor; taking <determinants> shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm1$, and the adjugate formula proves the converse. The <matrices> $\left(\begin{smallmatrix}1&m\\0&1\end{smallmatrix}\right)$ show infinitude.
Reduction modulo two is a homomorphism $G\to GL_2(\mathbb F_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.
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