The determinant map $\det\colon\operatorname{GL}(2,\mathbb R)\to\mathbb R^\ast$ defines a group homomorphism, so much is clear. This is just a reformulation of the fact that $\det(AB)=\det(A)\det(B)$. This map is also clearly surjective since
$$
\det\left(\begin{pmatrix}r&0\\0&1\end{pmatrix}\right)=r
$$
for any non-zero $r\in\mathbb R$. However, and here is the catch, this map is not injective; in fact, far from it. Note that for example $1\in\mathbb R^\ast$ has uncountably many pre-images given by
$$
\det\left(\begin{pmatrix}r&0\\0&r^{-1}\end{pmatrix}\right)=1
$$
for any non-zero $r\in\mathbb R$. And an isomorphism (the $\approx$ notation is usually reserved for isomorphisms) is by definition bijective. Hence, $\operatorname{GL}(2,\mathbb R)\not\approx\mathbb R^\ast$; well, at least not along this particular homomorphism.
The First Isomorphism Theorem tells you how to fix this issue: just mod out the kernel! The kernel is by definition given by $\operatorname{SL}(2,\mathbb R)$ as the identity element of $\mathbb R^\ast$ is $1$.
Regarding your question on the internal structure of the factor group.
Both $G=\operatorname{GL}(2,\mathbb R)$ and $K=\operatorname{SL}(2,\mathbb R)$ are mutliplicative groups. Given $A\in G$ its coset $AK$ is
$$
AK=\{AS\,|\,S\in K\}\,.
$$
Two elements in the factor group $G/K$ are the same if their representatives in $G$ differ by an element of $K$, i.e. $AK=BK$ iff $AB^{-1}\in K$. I think somewhere here you mixed up additive and multiplicative notions.
I don't have access to a copy of Gallian, so I cannot check, but the claim is false as stated.
Consider the group $G=S_3\times C_2$ of order $12=3\cdot2^2$, $3<2^2$. We see easily that $G$ has three subgroups of order four.
Best Answer
$(12)(13)(14) = \alpha(12)$ then multiply both sides by $(12)$ on the right, and $(12)(12) = e$, the identity, so
$\alpha = (12)(13)(14)(12) \in A_4$ (it has an even number of transpositions so it's in $A_4$, but the standard form is $(243)$ of course.
For the second we start from $$(1234)(12)(23) = \alpha(1234)$$ and multiply on the right by $(4321)$, the inverse of $(1234)$ and we get
$$\alpha = (1234)(12)(23)(4321) = (234) \in A_4$$