# A Quotient of the General Linear Group, Intuitively

/*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

*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.

*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!)