 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 $\theta (k_1{k}_2^{-1})=1_H$ for $k_1,k_2\in K$. Alsoso $gK{g}^{-1}\subseteq K$. 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 $\mathrm{im}\theta$. Thus the first isomorphism theorem givesThe quotient is isomorphic to all of $H$ precisely when the original homomorphism is surjective.

For the general linear group, the determinant gives a homomorphism $\det :\mathrm{GL}(n,\mathbb{R})\to {\mathbb{R}}^{\times }$, since $\det (AB)=\det A\det B$. Its kernel is the special linear group $\mathrm{SL}(n,\mathbb{R})$, so that subgroup is normal. Every nonzero real $t$ is the determinant of $\mathrm{diag}(t,1,\dots ,1)$ for $n\ge 1$, making the map surjective. Thereforewhere the operation on the right is multiplication.