Equivalence Relation definitions of Coset - looks like 1-step Subgroup Test? [Fraleigh p. 97 theorem 10.1]

Question

p. 4 We are especially interested in the case where the set is a group, and the equivalence relation has something to do with a given subgroup. That is, we want to partition a group G into subsets, each of which is determined by some fixed subgroup H. Once we have done this, we can write down a proof of Lagrange’s theorem in a nice way. Our present goal then is to find an equivalence relation on a group $G$ which is somehow related to a subgroup $H$. There are two very similar ones that we can define, and either one will work.

Let's return to our key example. The relation $\sim$ defined on $Z$ by $a\sim b\Leftrightarrow a\equiv b \mod n$
is equivalent to $n|(a-b)$ . This also means that $a-b\in nZ$. Let's try to generalize this.

Let $H\leq \text{ group }G$.
Define a relation $\sim_L$ on $G$ by: $a\sim_Lb$ if and only if $a^{-1}b\in H$.
Define a relation $\sim_R$ on $G$ by: $a\sim_Rb$ if and only if $ab^{-1}\in H$.
Then $\sim_L$ and $\sim_R$ are both equivalence relations on G.

(1.) Where do $a^{-1}b\in H$ and $ab^{-1}\in H$ loom from? Without writing them any other way, what do they mean? They look like the 1-Step Subgroup Test. Are they related?
Not querying where the coset definitions $aH = \{ah : h \in H\}$ and $Ha = \{ha : h \in H\}$ loom from. Nathan Carter discourses on this.

What are the equivalence classes of $\sim_R$?
Well, $[a]=\{x\in G: a\sim_R x\}=\{b\in G: \color{green}{ax^{-1}\in H}\}.$
I simplify the green separately, outside the set-builder notation:
$\color{green}{\iff}$ there exists $h\in H$ such that $ax^{-1}=h \quad \iff a\color{red}{h^{-1}}=x$.
Therefore the cell containing a is $\{ha : h \in H\}$, which we denote by $Ha$.

(2.) Because H is a subgroup, hence $\color{red}{h^{-1}} \in H$. Hence doesn't the paragraph overhead induce the left coset instead $ \{a\color{red}{h^{-1}} : \color{red}{h^{-1}} \in H\} = \{a\color{red}{\mathfrak{h}} : \color{red}{\mathfrak{h}} \in H\} $?

We generally use $Ha$ instead of the $[\cdot]$ notation when working in the group case. These equivalence classes actually have a special name, which we will use constantly from now on.

Answer 1

For the first part if we write the group additively we have $a-b$, while multiplicatively we have $ab^{-1}$. We could also have $-a+b$ (additive notation for groups often carries the assumption that they are commutative) which is written multiplicatively as $a^{-1}b$.

Then $n\mathbb Z$ is a subgroup of $\mathbb Z$. So the construction is the same, but with the group operation written multiplicatively.

Answer 2

$ab^{-1}$ is just the translation of $a-b$ (in the motivating example) from additive to multiplicative notation. And once you recall that the group operation is not necessarily commutative, you should consider using $b^{-1}a$ s an alternative case. Admittedly, this is not $a^{-1}b$, but the inverse of $b^{-1}a$.

You may want to check the exact use of variable names in the discussion of the equivalence classes of $\sim_R$. Some $b$ should be $x$ I assume.

And then there really seems to be some mixup. "$ax^{-1}=h\iff ah^{-1}=x$" is not correct. Instead, $$x\in [a]_R\iff a\sim_R x\iff x\sim_R a\iff xa^{-1} \in H\iff \exists h\in H\colon xa^{-1}=h\iff \exists h\in H\colon x= ha \iff x\in Ha$$

Answer 3

1. The idea comes from the analogy with integers: $$a \equiv b\ mod(n) \Leftrightarrow n | (a-b)$$ Or, in group notation $$ n|ab^{-1}$$

The set of multiples of $n$ is often denoted by $n\mathbb{Z}$ that give us:

$$a \equiv b\ mod(n) \Leftrightarrow ab^{-1} \in n\mathbb{Z}$$

2)

$$\{a\color{red}{h^{-1}} : \color{red}{h^{-1}} \in H\} = > \{a\color{red}{\mathfrak{h}} : \color{red}{\mathfrak{h}} \in H\}$$

Thats right! If $h \in H$ then $h^{-1} \in H$ so the two sets are the same.

Answer 4

I have always used this approach when teaching cosets and Lagrange's theorem. I use a different notation, though: \begin{gather} a\sim_H b\text{ means } ab^{-1}\in H \\ a\mathrel{{}_H{\sim}}b\text{ means } a^{-1}b\in H \\ \end{gather} in order to emphasize the role of $H$.

Where does this come from? Well, we want to define an equivalence relation using only the data, that is the group operation and the subgroup $H$. So, given two elements $a$ and $b$ it's natural to consider their product and asking whether it belongs to $H$. However, this is not a reflexive relation, but it becomes reflexive as soon as one of the two elements is inverted.

Note that reflexivity, with the definition of $\sim_H$, uses the fact that $1\in H$. Similarly, symmetry exploits the property that $x\in H$ implies $x^{-1}\in H$. Finally, transitivity uses the remaining property of subgroups.

What element should be inverted? It's immaterial: both ways define an equivalence relation. However the two relations may be different from each other. The fact that they coincide so reflects a special property of the subgroup: indeed the two relations coincide if and only if $H$ is a normal subgroup.

I like also to present the three properties in this order

1. $1\in H$ (no “variable” element involved)
2. if $x\in H$, then $h^{-1}\in H$ (one variable element involved)
3. if $x,y\in H$, then $xy\in H$ (two variable elements involved)

This makes clearer the connection with equivalence relations.

The equivalence class of $a\in G$ under $\sim_H$ is $$ [a]_{\sim_H}=\{x\in G: xa^{-1}\in H\}=Ha $$ that is, a translation of $H$ by $a$.