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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?

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  • $\begingroup$ 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 $\endgroup$
    – user829347
    Feb 14, 2022 at 22:53
  • $\begingroup$ 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. $\endgroup$
    – Alan
    Feb 14, 2022 at 23:25
  • $\begingroup$ "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. $\endgroup$ Feb 15, 2022 at 7:41

3 Answers 3

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$G$ 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 $H=\{1\}$.

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

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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.

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