A Quotient of the General Linear Group, Intuitively
Over the past few weeks, we've been chatting about quotient groups in hopes of answering the question, "What's a quotient group, really?" In short, we noted that the quotient of a group $G$ by a normal subgroup $N$ is a means of organizing the group elements according to how they fail---or don't fail---to satisfy the property required to belong to $N$. The key point was that there's only one way to belong to $N$, but generally there may be several ways to fail to belong. This, then, is why the interesting/substantial/meaty part of the quotient is contained in the set of nontrivial cosets. It's also why I like to think of $G/N$ as
"all the stuff in $G$ that's not in $N$. Sorta."
To make the "sorta" a little more concrete, we took a look at the quotient $\mathbb{Z}/5\mathbb{Z}$, just to get our feet wet. We then explored the intuition behind the First Isomorphism Theorem and what's inarguably one of the most useful quotients out there: that of a group by the kernel of a homomorphism. And last week, we used the First Isomorphism Theorem to get an intuitive feel for another example, namely the quotient of a group by its center.
Today I'd like to close out this series by looking at one last example. It's a quotient of a likely familiar group of matrices by a special subgroup.
Example #3: matrices and their determinants
Suppose $F$ is any field and $GL_n(F)$ is the group of invertible $n\times n$ matrices, a.k.a. the general linear group. Recall that "$SL_n(F)\subset GL_n(F)$" means "You belong to $SL_n(F)$ (the special linear group) if and only if you're an $n\times n$ invertible matrix with determinant 1." As you might expect, I think it's helpful to view $GL_n(F)/SL_n(F)$ as consisting of those matrices in $GL_n(F)$ with determinant not equal to 1.
To see this, first note that two matrices $A$ and $B$ live in the same coset if and only if $\text{det}(AB^{-1})=1$ which is true if and only if $\text{det}(A)=\text{det}(B)$. In other words, each coset in $GL_n(F)/SL_n(F)$ is simply the set of all matrices with a given determinant. So for example, if $F=\mathbb{R}$, then all invertible $n\times n$ matrices with $\text{det}=3$ are in the same coset, all those with $\text{det}=\sqrt{2}$ are in the same coset, and so on. In particular, if your determinant is not equal to 1, then it must be some other non-zero number. Intuitively, then, for any field $F$, we should expect the cosets of $GL_n(F)/SL_n(F)$ to be in one-to-one-correspondence with the all of the nonzero elements of $F$.
And our intuition is correct! If we define a map from $GL_n(F)$ to $F^\times=F\smallsetminus\{0\}$ by $A\mapsto\text{det}(A)$, we can check that it's surjective and has $SL_n(F)$ as its kernel. So by the First Isomorphism Theorem, the quotient $GL_n(F)/SL_n(F)$ must be isomorphic to $F^\times$. And this was exactly our observation: either you have determinant 1 or you don't, and there are exactly $|F^\times|\smallsetminus 1$ (this is larger than one, so long as $F^\times\neq\{1\}$) ways to fall into the latter category. Voila!
By the way, if you've ever wondered, "What does the determinant actually mean?", here's a nice thread on Math StackExchange that may shed some light on the situation. (There's a geometric answer!)