I'm continuing the discussion of the post: Identifying cosets of normal subgroups in $UT_2(\mathbb{R})$., where the topic is Artin's exercise 2.10.1:
Let $G$ be the group of invertible real upper triangular $2\times 2$ matrices. Determine whether or not the following conditions describe normal subgroups $H$ of $G$. If they do, use the First Isomorphism Theorem to identify the quotient group $G/H$. (a) $a_{11}=1$ (b) $a_{12}=0$ (c) $a_{11}=a_{22}$ (d) $a_{11}=a_{22}=1$.
In the FIT, there are 3 homomorphims: $\varphi : G \to G'$, $\pi: G \to G/H$ and $\overline{\varphi}: G/H \to G'$, where the first two are surjections and the latter is an isomorphism. I know that the kernel of $\pi$ is $H$. The post above mentions that in order to identify the quotient groups $G/H$, I need to find a surjection $\varphi: G \to G'$ whose kernel is $H$. But isn't this the wrong way to go about this because $H$ is the kernel of $\pi$ not $\varphi$?
Second question: In identifying the cosets, what exactly is the equivalence condition for two upper triangular matrices to be in the same coset?
Best Answer
The way I understand the First Isomorphism Theorem is the following:
(That means that $f=\overline{f}\circ\pi$, where $\pi\colon G\to G/N$ is the canonical surjection, giving the "three maps" you mention.)
So the best way to understand the First Isomorphism Theorem, in my opinion, is that it starts with $f$ and $f$ alone. Then you use $f$ to determine $N=\ker(f)$, and to define the morphism $\overline{f}\colon G/N\to G'$. It tells you that $\overline{f}$ is one-to-one and a homomorphism, with the same image as $f$.
(You may assume $f$ is given as a surjection; in that case, $\mathrm{Im}(f)=G'$; the way I phrased it is just a little more general in set-up.)
Now, sometimes you are in a different situation. Sometimes you have a group $G$ and a normal subgroup $N$, and you want to know whether the group $G/N$ has some other "more common" name and presentation. For example, if $G=S_3$ and $N=\langle (123)\rangle$, it is much more useful to know that $G/N$ is isomorphic to "the cyclic group of order $2$" than to deal with the cosets directly. That is the situation you (potentially) have here: you have your group $G$, and in some of the cases a normal subgroup $H$; you know you can construct the quotient group $G/H$ whose elements are cosets, and multiply them via representatives, etc. But in the case of this group, it may not be clear if that group has a "nicer"/"cleaner" presentation. In that situation, if $\mathfrak{G}$ is a "nicer" name for the group (as you noticed in the previous post, for example, that in some cases it is isomorphic to $\mathbb{R}^{\times}$), then the way to use the First Isomorphism Theorem to "realize" that identification is to find a surjective morphism $f\colon G \to \mathfrak{G}$ whose kernel is the normal subgroup $H$ you have in mind.
For instance, in the case of $S_3$ and $\langle (123)\rangle$, I might define a map $s\colon S_3\to \{\pm 1\}$ by "$s(\sigma)$ is $1$ if $\sigma$ is even, and $-1$ if $\sigma$ is odd." Then verify this is a homomorphism and that the kernel is precisely the subgroup I have in mind, $\langle (123)\rangle$; those two verifications establish that $S_3/\langle (123)\rangle \cong \{\pm 1\}$. This is what one means by saying you use the First Isomorphism Theorem to "identify" or "recognize" the quotient as a more familiar group.
As to you second question... for any group $G$ and subgroup $K$, we know that two elements $x,y\in G$ are in the same (left) coset of $K$ if and only if $xK=yK$, if and only if $y^{-1}x\in K$, if and only if $x^{-1}y\in K$. So you could check that condition to test whether two diagonal matrices are in the same coset relative to a specific $H$... but I'm not sure why you would want to. You don't want to work with the cosets directly: that's the point of the First Isomorphism Theorem. You want to work with the whole group, and define a morphism with specific kernel so that you don't have to work with the cosets and cosets representatives, and instead can work with the target group.