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Suppose I am given an equivalence relation on set A which is endowed with addition & closure under addition.

Then, to have addition on equivalence classes of A to be well-defined, it is to show:

$[a]+[b] = [a+b]$

Suppose I have shown that if $\forall a,b \in A, a \sim a', b\sim b'$ then $a+b \sim a'+b'$, then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$.

But then, now I need to show the other direction. I found it is easiest to define $c \in A$ so $b+c = 0$. So if $x \in [a+b]$ then $x \sim a+b$, and using what I proved, $x+c \sim a+(b+c) = a$. Letting $r = x-b$ and $s = b$ shows the other direction.

My question now is, to do this, (A,+) seems to need to have the property that all elements must have an identity, an inverse, and is associative, so what exactly must (A,+) be for all this to work, a group? abelian group? and why?

Note: Also I think I'm failing to consider a lot of things when thinking about this, so it will be very helpful if anyone could point many of those things out.

nabu1227
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3 Answers3

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$(A,+)$ does not need to be a group. For example, we may take $A=\Bbb N$ (with usual addition) and define $a\sim b$ as "$a$ and $b$ end in the same decimal digit". Then $(A/{\sim},+)$ is a group (cyclic of order $10$).

Similary, $(A/{\sim},+)$ may be an abelian group even if $(A,+)$ is non-abelian. (But not vice versa).

Hagen von Eitzen
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  • Thanks, but if A is a group, then this will definitely work? Also, are associativity and existence of identity and inverse (not necessarily unique) a sufficient condition? Have I also added an extra sufficient condition? – nabu1227 Jan 10 '21 at 13:29
  • @LHC2012 If $A$ is a group (here written abelien with $+$, but that's not needed) and $[a]+[b]:=[a+b]$ is well-defined, then $+$ on $A/{\sim}$ is associative because $[a]+([b]+[c])=[a]+[b+c]=[a+(b+c)]$ etc. Also, $[0]$ is neutral, and $[a]+[-a] = [a+(-a)]=[0]$, i.e., $[-a]$ is inverse to $[a]$. So $A/{\sim}$ is a group. -- In that case, also $H:=[0]$ is a subgroup of $A$! Namely, $a\sim 0$ and $b\sim0$ implies $a+b\sim 0$ and $-a\sim 0$. Moreover, $H$ is a normal subgroup of $A$ (trivially forabelian groups, but also in the general case), and it turns out that $A/{\sim}\cong A/H$ canonically. – Hagen von Eitzen Jan 10 '21 at 16:49
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If $A$ is a set, $+$ is a binary operation on $A$ and $\sim$ denotes an equivalence relation on $A$ then we can define a binary operation on $A/\sim$ by means of:$$[a]+[b]:=[a+b]\tag1$$if and only if $$\forall a,a',b,b'\in A [a\sim a'\wedge b\sim b'\implies a+b=a'+b']\tag2$$

"...then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$..."

No, $(1)$ tells us directly that $[a+b]=[a]+[b]$ and there is no such thing needed as "..showing the other direction..".

The only thing you really need is a proof that $(2)$ is valid. It is actually the proof that the operation "defined" in $(1)$ is indeed well defined.

drhab
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  • Why is (2) equivalent to (1), can you elaborate on that? Thanks! – nabu1227 Jan 11 '21 at 01:05
  • What I am really asking is why (2) accounts for the other direction as well. – nabu1227 Jan 11 '21 at 03:06
  • It is not stated that $(1)$ and $(2)$ are equivalent. It is stated that $(1)$ is a "well" definition iff condition $(2)$ is satisfied. Essential is that $[a]+[b]$ is defined by use of the represetatives $a$ and $b$. "Ho, wait!.." we say then. "How can we be sure that this does not depend on those chosen representatives? Of course the outcome must not be different if we choose another pair of representatives." Well, $(2)$ is exactly the assurance that this will not happen and makes us say that the operation is well-defined. – drhab Jan 11 '21 at 08:57
  • We are dealing with a definition here and a definition $a:=b$ does not need any proof of proof of form $a\subseteq b$ and conversely $b\subseteq a$. However what the definition does need here is a checkup that it is well formulated. This necessary and sufficient checkup is $(2)$ in my answer. – drhab Jan 11 '21 at 09:14
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So you have a semigroup (a single binary operation that is associative on your set $A$). The magic words are “congruence relation”, which are exactly what drhab gives in his answer. You can see a lot of this at play in this previous answer about groups.

Definition. Let $S$ be a semigroup. A congruence on $S$ is an equivalence relation $\sim$ on $S$ such that for all $x,y,x’,y’\in S$, if $x\sim x’$ and $y\sim y’$, then $xy\sim x’y’$.

Proposition. An equivalence relation on $S$ is a congruence if and only if, when considered a subset of $S\times S$, it is a subsemigroup of the latter under coordinatewise operation.

Proof. If $\sim$ is a congruence, then by definition the subset is closed under the coordinatewise operation, hence is a subsemigroup. Conversely, if an equivalence relation is a subsemigroup, then $x\sim x’$ and $y\sim y’$ mean $(x,x’),(y,y’)\in\sim$, and by closure we get $(xy,x’y’)=(x,x’)(y,y’)\in \sim$, hence $xy\sim x’y’$, hence $\sim$ is a congruence. $\Box$

Given an equivalence relation $\sim$ on $S$, let $S/\sim$ be the set of equivalence classes, and let $[s]$ denote the equivalence class of $s\in S$ under $\sim$.

Proposition. If $\sim$ is a congruence on $S$, then $S/\sim$ is a semigroup under the operation $[s][t]=[st]$.

Proof. The operation is well defined: if $[s]=[s’]$ and $[t]=[t’]$, then $s\sim s’$ and $t\sim t’$, hence $st\sim s’t’$, so $[st]=[s’t’]$. The operation is associative, as $$([s][t])([v]) = [st][v] = [(st)v] = [s(tv)] = [s][tv] = [s]([t][v]).$$ Thus, $S/\sim$ is a semigroup. $\Box$

And conversely:

Theorem. Let $S$ be a semigroup, and let $\sim$ be an equivalence relation on $S$. Then the operation on $S/\sim$ given by $[s][t]=[st]$ is well-defined if and only if $\sim$ is a congruence on $S$.

Proof. That the operation is well-defined when $\sim$ is a congruence is given above. Conversely, assume the operation is well defined: that means that if $s\sim s’$ and $t\sim t’$, then $[st]=[s][t] = [s’][t]’] = [s’t’]$, hence $st\sim s’t’$, proving that $\sim$ is a congruence on $S$. $\Box$

Congruences play the role of normal subgroups for groups and two-sided ideals for rings in the context of general algebras, the study of Universal Algebra. The isomorphism theorems have counterparts for general algebras using congruences.

Thus, you don’t need this to be a monoid, or a group, for this to work; you just need a subsemigroup.

Arturo Magidin
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