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I want to get an intuitive idea of the operation being "well-defined" for quotient groups. So, let's say I have a group $G$, with subgroup $H$, and let's say my set of left cosets is $G/K$. My lecture note says this: If $H$ is normal, then $G/H$ is a group under binary operation $aH \circ bH = (ab)H$. So let's say I am asked whether $G/H$ forms a group where $H$ is not normal, and I have already determined that. My professor referred to the following which I am not sure if I totally understand:

Show that there are $a, b, a', b' \in G$ such that $ aH = a'H$ and $bH = b'H$ but $aH \circ bH = (ab)H \neq a'H \circ b'H = (a'b')H$, and you are done, and I did what she suggested, but I am not sure what is going on. So,

$1.$ What did I exactly show by proving what my professor suggested?

$2.$ In general, is it a strategy that every time you have to prove $G/H$ does not form a group for a non-normal $H$, you show that the operation is not well-defined?

Also, a quick google search also showed me that the theorem that talks about $G/K$ forming a group is an "if and only if" statement and not difficult to prove as well. Still, any help on my questions above would be great.

Four Seasons
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    For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose – J. W. Tanner Apr 10 '19 at 00:56
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    If $C,D$ are cosets of $H$, we'd like to define $CD$ by taking a random element of $a \in C$ and a random element of $b \in D$ and define $CD = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result. – Jair Taylor Apr 10 '19 at 01:17

3 Answers3

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In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.

In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.

When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets $(2\mathbb{Z} + 1) \circ (2\mathbb{Z} + 1)$ is $2\mathbb{Z}$.

There is an alternative definition. You can define the product of two cosets $A$ and $B$ as $$ A \circ B = \{ ab \ | \ a \in A \text{ and } b \in B \}. $$ This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.

Ethan Bolker
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  • I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation? – Four Seasons Apr 10 '19 at 01:11
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    Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition. – Ethan Bolker Apr 10 '19 at 01:17
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Perhaps a concrete example will make it clearer?

We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.

We need a non-normal subgroup. Let's take $H=\{e,(12)\}$.

We need two cosets. Using $H$ itself might be too trivial, so it seems most promising to take $$ a=(23) \qquad aH = \{(23),(132)\} = a'H \qquad a'=(132) $$ $$ b=(13) \qquad bH = \{(13),(123)\} = b'H \qquad b'=(123) $$

Now, if we had a quotient group what should the product $\{(23),(132)\}\circ\{(13),(123)\}$ be?

From one perspective we have $$(23)H\circ(13)H =^? (123)H$$ But we could also say $$(132)H\circ(123)H =^? eH = H$$

But the product of the coset $\{(12),(132)\}$ with the coset $\{(23),(123)\}$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say it should be two different things! So we're in trouble.

hmakholm left over Monica
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Here's a different description that gives a different lens on all of this.

An equivalence relation on a set is a binary relation (call it $\sim$) that is reflexive ($x\sim x$), symmetric ($x\sim y\implies y\sim x$), and transitive ($x\sim y \land y\sim z \implies x\sim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $a\sim b \iff ab^{-1}\in H$. Prove that this is indeed an equivalence relation.

An equivalence class of an element $x$ with respect to some equivalence relation $\sim$ is the set $\{y\mid x\sim y\}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]=\{g\mid 1\sim g\}=\{g\mid g^{-1}\in H\}=H$.

Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $x\sim y$ as $x$ "equals" $y$, and so if $x\sim x'$ and $y\sim y'$ we would want $xy\sim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY=\{xy\mid x\in X\land y\in Y\}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $\sim$" is the same as "$\sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.

So the first question is: is the equivalence defined above a congruence?

Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $\approx$. We have $a\approx b\iff ab^{-1}\approx 1$. (Why?) We can thus show $a\approx b\iff ab^{-1}\in[1]$, mimicking our first equivalence relation.

Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $\sim$?

Derek Elkins left SE
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