= Solution
The <kernel of a group homomorphism> is $K=\{g:\theta(g)=1_H\}$. It is a <subgroup>, since the identity maps to the identity and $\theta(k_1k_2^{-1})=1_H$ for $k_1,k_2\in K$. Also
$$
\theta(gkg^{-1})=\theta(g)\theta(k)\theta(g)^{-1}=1_H,
$$
so $gKg^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, proving that $K$ is a <normal subgroup>.
The map $gK\mapsto\theta(g)$ is well defined, multiplicative and injective, with image $\operatorname{im}\theta$. Thus the <first isomorphism theorem> gives
$$
\boxed{G/K\cong\operatorname{im}\theta\leq H.}
$$
The quotient is isomorphic to all of $H$ precisely when the original homomorphism is surjective.
For the <general linear group>, the <determinant> gives a homomorphism $\det:\operatorname{GL}(n,\mathbb R)\to\mathbb R^\times$, since $\det(AB)=\det A\det B$. Its kernel is the <special linear group> $\operatorname{SL}(n,\mathbb R)$, so that <subgroup> is normal. Every nonzero real $t$ is the determinant of $\operatorname{diag}(t,1,\ldots,1)$ for $n\geq1$, making the map surjective. Therefore
$$
\boxed{\operatorname{GL}(n,\mathbb R)/\operatorname{SL}(n,\mathbb R)\cong\mathbb R^\times,}
$$
where the operation on the right is multiplication.
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