# Multiplying cosets is well-defined

Let $$(G, *)$$ be a group and $$(H, *)$$ its normal subgroup. Then the factor group is defined as

$$G/H := \left\{g*H: g\in G \right\} .$$

We define the multiplication of cosets (elements of G/H) as a function $$f: G/H \times G/H \rightarrow G/H$$given by $$\left(g_1 H, g_2 H \right) \mapsto (g_1 * g_2)*H$$. Usually, this is shown to be well-defined function by using the "definition":

A map $$f: A\rightarrow B$$ is well-defined if for every $$a, b \in A$$ $$a=b$$ implies $$f(a)=f(b)$$.

However, this "definition" does not make sense to me: If $$a=b$$, doesn't $$f(a)=f(b)$$ always follow, since I can always rewrite $$a$$ as $$b$$?

That is why I am hoping to find a more "explicitly" defined function which defines the same multiplication of cosets. To do this, I was trying to find a function $$\tilde{f}: G/H \rightarrow G$$ which maps every $$g*H \in G/H$$ satisfying $$\hat{g}*H=g*H$$ to $$\hat{g}$$. After that, I could define another function $$\overline{f}: G \rightarrow G/H$$ given by $$g \mapsto g*H$$. Then I redefine the function $$f$$ as $$(x, y) \mapsto \overline{f}(\tilde{f}(x)*\tilde{f}(y))$$, which is clearly well-defined.

Now the problem is, can such function $$\tilde{f}$$ be found?

• It is implicitly assumed that when one talks of a 'function' $f:A\to B$, they mean a well-defined one. After all, it would be erroneous to refer to it as a function if it made no sense. So yes, $a=b\implies f(a)=f(b)$ is automatic Feb 14, 2022 at 22:53
• The issue arises when you define a function on a set/equivalence class/etc. by how it acts on one member. You have to show that no matter the choice, you get the same output.
– Alan
Feb 14, 2022 at 23:25
• "To do this, I was trying to find a function f~:G/H→G which maps every g∗H∈G/H satisfying $\hat{g}$∗H=g∗H to $\hat{g}$." I'm not clear on what you're saying. What is $\hat{g}$? Given some property $P$ that you want $f$ to satisfy, there's a big difference between $\exists f:\forall \hat{g} P(f)$ versus $\forall \hat{g} \exists f:P(f)$, and it's not clear which you're talking about. Feb 15, 2022 at 7:41

$$G$$ is divided into disjoint subsets $$gH$$. You can pick one element $$\tilde g$$ in every coset $$gH$$ (which can be done using AC). Then define $$\tilde f(gH)$$ as $$\tilde g$$. This can be done in many ways unless $$H=\{1\}$$.

• Can you clarify what is meant by AC? Feb 14, 2022 at 23:43
• Axiom of choice. Feb 14, 2022 at 23:48
• Thank you! This is exactly what I was hoping for! Feb 14, 2022 at 23:54

The point of the definition which you quote can be seen from the following:-

Suppose you defined a map from the rationals by $$f(\frac{a}{b})\rightarrow a$$.

You would also have $$f(\frac{2a}{2b})\rightarrow 2a$$.

Since $$\frac{a}{b}=\frac{2a}{2b}$$ we see that $$f$$ is not well-defined.

If a=b, doesn't f(a)=f(b) always follow, since I can always rewrite a as b?

That rewriting $$a$$ as $$b$$ results in the same output for $$f$$ is exactly what is under concern. If might be clearly if it's worded as "If two expressions represent the same object, then $$f$$ applied to each expression gives the same result". See, the issue is that the definition of $$f$$ that you quote defines the output in terms of a particular representation of elements. So the question is, if $$g_1$$ and $$g_1'$$ define the same coset, and $$g_2$$ and $$g_2'$$ another coset, do $$g_1g2$$ and $$g_1'g_2$$ define the same coset?

As for your idea of a more explicitly defined function, besides requiring the Axiom of Choice, this has much the same problems as the original definition. While it avoids the problem of not being well-defined, the cases where the original would not be well defined also are cases where your definition would be problematic. For instance, suppose I want to take $$(g_1H)(g_2H)(g_3H)$$. According to the property of associativity, this is well-defined; it doesn't matter if we interpret as $$(g_1H)((g_2H)(g_3H))$$ or $$((g_1H)(g_2H))(g_3H)$$. But does your definition yield the same result either way? It's possible to prove that if $$H$$ is normal, then it does, but that proof isn't any easier than just proving that the original definition is well-defined.