I'm reading a textbook (Contemporary Abstract Algebra, 9th Edition, Joseph A. Gallian), and it gave the following example:
Recall that $SL(2,\mathbb{R})=\{A\in GL(2,\mathbb{R}\ |\det A=1)\}$ and let $H=\{A\in GL(2,\mathbb{R}\ |\det A=\pm 1)\}$. Then mapping $\Phi(A)=\det A$ fom $GL(2,\mathbb{R})$ onto $\mathbb{R}^*$ shows that $GL(2,\mathbb{R})/SL(2,\mathbb{R})\approx\mathbb{R}^*$.
Here, $\mathbb{R}^*$ is used to represent the set of all real nonzero numbers.
Where I'm getting tripped is why $GL(2,\mathbb{R})$ itself isn't simply homomorphic to $\mathbb{R}^*$. By definition, $A\in GL(2,\mathbb{R})$ implies that $\det A\neq0$, and as we all know, $\det AB=\det A\det B$. This to me seems to imply that $GL(2,\mathbb{R})\approx \mathbb{R}^*$. How can a factor group and the group it is a subgroup of be homomorphic to the same group?
(Additionally, I think I'm still a bit fuzzy as to what $GL(2,\mathbb{R})/SL(2,\mathbb{R})$ actually is. To the best of my knowledge, I think it's the set of all cosets of $SL(2,\mathbb{R})$ in $GL(2,\mathbb{R})$ – that is, all the matrices $C$ where $AB=A'B'+C$, with $A,A'\in GL(2,\mathbb{R})$ and $B,B'\in SL(2,\mathbb{R})$. Is this an accurate way of thinking about the group?)
Best Answer
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.