Understanding the usage of being "well defined" function in Orbit Stabilizer Theorem

Question

I am first year university student, we saw some proofs in class. I am trying to understand them from other sources if I could not understand them at that moment, because the class is crowded and I am ashamed of asking stupid question among others.

The following question is one of them. I am trying to understand the proof of orbit stabilizer theorem, but I stuck in one point

I made research about the proof of orbit-stabilizer theorem. I have seen some proof approaches, but something was interesting for me. The interesting part was showing well-define property of bijection function. Let me introduce you a part of a sample proof:

We define a one-to-one map $ψ$ from $Gx$ onto the collection of left cosets of $G_x$ in $G$ such that $\psi :g(x) \rightarrow gG_x$. We must show that this map ψ is well defined,so show that if $g(x)=h(x)$ where $g,h \in G$, then $gG_x=hG_x$.

Here $Gx$ denotes orbits and $G_x$ denotes stabilizer

I did not write the rest of proof,because my question is up to now. I saw some proofs in my other math classes such that we used bijective functions for showing one to one correspondence between sets to show that their cardinalities are equal, but we have never needed to prove that the functions we selected for bijectivity is well defined by using the condition if $x=y$ where x and y are elements of domain,then $f(x)=f(y)$

Look here for well defined

However, when I look at all proofs in books and internet, they always firstly prove that the bijection function $\psi :g(x) \rightarrow gG_x$ is well defined. Then, why did they need to show its well-defined property in this proof. If showing firstly being well-defined function is mandatory in bijective functions, why dont we always see it in some other proofs ?

Can you please enlighten me the reason why being well defines is always proved in all resources ?

Some resources I made use of :

• Contemporary Abstract algebra page $147$

• A first course in Abstract ALgebra page $158$

Answer 1

Proving well-definedness is a requirement when working with equivalence classes.

Let $\sim$ be an equivalence on $X$. Then well-definedness of a function $f:X/\sim\to Y$ is when

$$[x]_\sim=[y]_\sim\implies f([x]_\sim)=f([y]_\sim);$$

that is, the image of $f$ does not depend on the representative $x\in [x]_\sim$, where $[z]_\sim$ is the $\sim$-equivalence class of $z$.

This does not require $f$ to be bijective.

Your $G_x$ are equivalence classes.

---

Consider an illustrative example of when it fails: define

$$\begin{align} f:\Bbb Q&\to \Bbb Z,\\ \frac{a}{b}&\mapsto ab. \end{align}$$

We have $\frac{1}{2}=\frac{2}{4}$ but

$$\begin{align} f\left(\frac12\right)&=1\times 2\\ &=2\\ &\neq 8\\ &=2\times 4\\ &=f\left(\frac24\right). \end{align}$$

Answer 2

General remarks.

Just writing down a definition does not mean that the definition is sensible; that is, you cannot define something into existence.

Often, when working with functions, we provide an explanation of what the function is supposed to do. But it is not necessarily evident that what we have described will actually yield (i) a function, that (ii) is defined on the entire putative domain, and (iii) will take values in the putative codomain. When any of those items is unclear, we usually engage in a verification that our description does, in fact, yield a function with that domain and that codomain. This verification is called "checking that the function is well-defined."

Note that "well-defined"ness is not a property that is defined in a specific way. (As I have said before, "begin well-defined is not a notion that is well defined").

How can (i) fail? It can fail if our description is such that it is possible for the same input to have more than one output. Remember that a function is supposed to take each input to one and only one output. Say we tried to define a function from the set of human beings that have a phone to the set digits, and said "Map each person that has a phone to the first digit of their phone number." The problem is that the same person may have multiple numbers, which may have multiple different first digits. Then your map is not a function, because the same person may be mapped to several different outputs.

How can (ii) fail? Maybe your definition does not apply to every element of the domain. If you say "Let $f$ be a function with domain all real numbers and codomain all real numbers, which sends the number $x$ to its reciprocal $\frac{1}{x}$." That does not give you a function from $\mathbb{R}$ to $\mathbb{R}$, because there are elements in the domain on which this rule does not work (namely, $x=0$). Or try to define a function from all people currently alive to digits by sending each person to the first digit of their primary cell-phone number. Then your function is not properly defined because there are people currenlty alive that have no cell-phone number at all.

How can (iii) fail? It can fail if your rule gives a value, but that value is not in your described range. For example, try to define a function from the set of all human beings currently alive to the latin alphabet, by sending each person to the first letter or logogram/ideogram of their legal name. This has the problem that there are people whose legal name is not in the latin alphabet, and as such the map does not take values in your desired set.

And sometimes a function can have issues in all three items.

Specific remarks.

In algebra, when we talk about functions being "well-defined", the situation is usually an issue with (i) above, but in a very specific circumstance:

We have a set $X$ and an equivalence relation on $X$, say $\sim$. We want to define a function from the set of equivalence classes $X/\sim$ to some set $Y$. Each equivalence class is the class of someone, $[x]_{\sim}=\{a\in X\mid a\sim x\}$. We then give a definition which explains the value we want $f$ to take on $[x]_{\sim}$ in terms of $x$.

The problem here is that there may be $x\neq y$ in $X$ with $[x]_{\sim}=[y]_{\sim}$. Then we need to verify that the description of the value in terms of $x$ will yield the same final result as the description of the value in terms of $y$.

For example, suppose you have the integers as your set, and $\sim$ is the relation "have the same parity". So your set $X/\sim$ is the set consisting of two elements: one is "odd integers" and the other is "even integers." "Odd integers" is the set $[x]_{\sim}$ for any odd integer $x$. So: $$\text{odd integers} = [1]_{\sim} = [-1]_{\sim} = [3]_{\sim}=[-3]_{\sim} = \cdots$$ and "even integers" is the set $[y]_{\sim}$ for any even integer $y$. So: $$\text{even integers} = [0]_{\sim} = [2]_{\sim} = [-2]_{\sim} = \cdots$$ Then I want to define a function from $X/\sim$ to the natural numbers by saying "send $[x]_{\sim}$ to $|x|$."

This is problem, because even though $[1]_{\sim}$ is exactly the same element as $[3]_{\sim}$, my instructions tell me that I need to send $[1]_{\sim}$ to $1$ and $[3]_{\sim}$ to $3$. So this definition does not give me a "well defined function" because the same element of the domain is mapped to multiple elements.

This is the sitation you find yourself in. You have two sets:

1. The orbit of $x$ under the action of $G$, $Gx = \{gx\mid g\in G\}$.
2. The set of left cosets of the stabilizer of $x$ in $G$, $gG_x$, where $G_x=\{g\in G\mid gx=x\}$.

The elements of $Gx$ are "called" $gx$ with $g\in G$. But the same element may have multiple names, because you can have $gx=hx$ with $g\neq h$.

The elements of the set of left cosets are "called" $gG_x$; but the same left coset may have multiple names, because you can have $gG_x = hG_x$ with $g\neq h$.

The function we are trying to define says:

1. Take an element $y$ in $Gx$.
2. Then we know that there exists $g\in G$ such that $gx=y$. Pick such a $g$.
3. Define $f(y)$ to be $gG_x$.

The problem here, then, is item (i) mentioned above: what happens if you pick a different element $h$ such that $hx=y$? The elements of $Gx$ may be obtained in multiple ways (that is, they have "multiple names"), but you are describing the image in terms of the specific name you have chosen. What if you choose $g$, but I choose a different element $h$ such that $hx=y$? Then when you read the instructions of $f$ you get the coset $gG_x$; and when I read the instructions I get the coset $hG_x$. They look different. That's a potential problem.

But then we remember that here, the elements of the codomain also have multiple names. So it is possible for $gG_x$ to equal $hG_x$ even if $g\neq h$. So it is possible that we don't actually have a problem. This is where checking that our instructions give a "well-defined function" come in.

What we need to show is that if we choose different "names" for the input (if we make different choices in step 2), then what we get will be different "names" for the same out put. In other words, we need to verify that

If $gx = hx$, then $gG_x=hG_x$

with an unspoken "even if $g\neq h$".

We need to make sure this works before we start checking if $f$ is bijective! Because if what we have described is not even a function, then checking whether it is bijective is nonsense. It would be like checking that the cheese that the Moon is made out of Stilton and not Cheddar. Garbage in, garbage out. Until we know there is actually a function here, we can't even begin to check whether it is bijective or not.

This happens a lot in algebra (and other situations) because we often have functions defined on equivalence classes, and we define them in terms of representatives for those equivalence classes. It doesn't happen as often in things like Combinatorics or Set Theory when we are trying to verify that two sets have the same cardinality because usually those sets are not defined in terms of equivalence classes, or the functions we are defining are not defined in terms of representatives. That's why you have probably not seen this issue of "well-defined"ness when looking at proofs that certain sets have the same cardinality, but why you will see it very often in abstract algebra.