# I don't understand equivalence relations

I'm trying to show that the left coset with a subgroup is an equivalence relation. So taking some element $g \in G$ and a subgroup $H$, the left coset is defined as $gH = \{gh : h \in H\}$. That means $g$ is fixed for this subgroup, as far as I can tell from what I'm reading.

So first, reflexivity: We have to show that $h$ can be mapped to itself under the definition of the left coset. So we have $gh = h$, but since $g$ is fixed and we cannot choose $g$, I don't know what to do here. I know for a fact that the left coset with a subgroup is an equivalence partition, but out of the dozen equivalence relation proofs I've read for this, I don't understand a single one of them. For example, this one site takes a group $G$ and subgroup $H$, then defines $a = bh$ and aims to prove the relation $a \sim b$. Why do you have to define $a = bh$, and why are we proving the relation $a \sim b$? I thought we need to prove the relation coset+partition $\sim$ group?

-
The whole set of equivalence classes (i.e., all the left cosets) is reflexive, symmetric and transitive, not one single class, @user130018 ! –  DonAntonio Mar 21 '14 at 0:19
I'm not sure what equivalence classes are, but what do you mean by all the left cosets? Do you mean it is $GH = gh$ for all $g \in G$ and $h \in H$? –  ᴇʏᴇs Mar 21 '14 at 0:21
I think you better study equivalence classes (google it), and then you get into left cosets of some subgroup in a group, otherwise you may get seriously confused... –  DonAntonio Mar 21 '14 at 0:22
I tried to read some pages I searched for on Google, but I don't understand any of it. Do you have a website that makes it very easy for a dumb person to understand? –  ᴇʏᴇs Mar 21 '14 at 0:24
Consider all arithmetic progressions of fixed common difference. Allow arithmetic progressions to begrowing both directions (that is, no first term). Example for $d=4$: $A_0=\{\ldots, -8,-4,0,4,8,\ldots\}, A_1=\{\ldots, -7,-3,1,5,9,\ldots\}, A_2=\{\ldots, -6,-2,2,6,10,\ldots\}. A_3=\{\ldots, -5,-1,3,7,11,\ldots\}$. The first one is a subgroup and together these 4 sets form the collection of all cosets. –  P Vanchinathan Mar 21 '14 at 0:31

The thing you may be missing is that equivalence relations and partitions are the same thing, in the sense that for every equivalence relation there is a natural partition, and vice versa.

Suppose you have a partition of some set $S$ into parts. For concreteness, let's say that the set $S$ is the people in Europe, and the parts are the populations of European countries: All the people in France, all the people in Spain, and so on. The rule for a partition is that every person must be in exactly one part, and this division into countries is a partition, because each person in Europe is in exactly one country.

The corresponding equivalence relation, is that person $b$ is equivalent to person $a$ if $b$ is in the same part of the partition that $a$ is in. That is, if $b$ is in the same country as $a$. We will write this as $a\sim b$. This relation $\sim$ is an equivalence relation:

• Each person $a$ is in the same country as him or herself, so for each $a$ we have $a\sim a$, and the relation is reflexive
• If $a \sim b$, then $a$ is in the same country as $b$, so $b$ is in the same country as $a$, so $b\sim a$, and the relation is symmetric
• If $a\sim b$, and $b\sim c$, then $a$ is in the same country as $b$, and $b$ is in the same country as $c$, so $a$ is in the same country as $c$, so $a\sim c$, and the relation is transitive.

For any partition, there is an equivalence relation that says the same thing, namely the relation of being-in-the-same-part-as.

Working the other way, if we have any equivalence relation on some set, say $\sim$, then there is a corresponding partition of that set: for each element $x$, we can consider the set of all the $y$ that are related to $x$. For the equivalence relation of "being-in-the-same-country-as", we take, for each person, say Angela Merkel, there is a part of the partition which is the set of all the people in the same country as Angela Merkel. If we agree to write $[p]$ to mean the set of all people in the same country as person $p$, then $[\text{Angela Merkel}]$ is the set of all the people in the same country as Angela Merkel.

This is a partition: Certainly each person $p$ is in at least one of the parts, namely $[p]$. And it also turns out that no person is in two different parts. For suppose we knew that $p$ was in part $[x]$ and also in part $[y]$. Then $p$ is in the same country as person $x$, and also in the same country as person $[y]$. But then $x$ and $y$ are in the same country as each other, and $[x]$ and $[y]$ are not two different parts but the same part, so $p$ is only in one part, not two parts.

For any equivalence relation, there is a partition that says the same thing, namely the one where the parts consist of groups of related elements.

So partitions and equivalence relations are two different ways of looking at the same thing.

If you want to show that some relation is an equivalence relation, one way to do that is by showing that it partitions its domain, that each element is in exactly one part.

Or if you have a family of sets, and you want to show that it is a partition, you can define a relation $\sim$, where $a\sim b$ means that $b$ is in the set belonging to $a$, and show that it is an equivalence relation. That is what we are doing here.

Now how does this relate to left cosets? For each element $g$ in the original group, we have the “left coset of $g$ by $H$”, which means the set of everything you can get by taking some element $h$ from $H$, and computing $g\cdot h$. Each $g$ has its own coset, which we will write as $gH$. We want to show that the cosets partition $G$.

Since partitions and equivalence relations are the same thing, we can do this by showing that a certain relation, $\sim$, is an equivalence relation. This might be easier than showing directly that the cosets partition $G$. The relation $\sim$ we need to consider is the one that says that $g\sim g'$ if $g'$ is in $g$'s coset. (This is just like the example before, where we had $p\sim p'$ if $p'$ was in $p$'s country.) So we have this $\sim$, and we want to show it is an equivalence relation. If it is, we know automatically that the cosets for a partition of $G$, because equivalence relations and partitions are the same thing.

I hope this makes more sense.

-
Sorry, I think I may have worded my question incorrectly. I wanted to prove that the left cosets partition a group, and it said something about equivalence relations so I'm trying to understand it. Does having an equivalence relation of two elements being in the same coset help with my goal at all? –  ᴇʏᴇs Mar 21 '14 at 0:41
Sorry, I will write an addendum ahortly. –  MJD Mar 21 '14 at 0:56
The part I don't understand about equivalence relations is that you say it describes how "person $b$ is equivalent to person $a$ if $b$ is in the same part of the partition that $a$ is in", but how does this show that all the countries make up Europe? To me, all this "equivalence relation" says is that one person is in the same country as another person, and says nothing about all the countries making up the continent. So in concrete English, equivalence relation says to me that person A and person B are in England. And nothing about England, France, Germany, etc. making up Europe. –  ᴇʏᴇs Mar 21 '14 at 12:25
I'm not talking about the countries, I'm talking about the populations of the countries. The parts aren't England, France, Germany, but rather the populations of England, France, Germany. And how do we know that these separate populations make up the population of Europe? Well, suppose the contrary, suppose there was some person $p$ that was in Europe but not in one of those populations. But $p$ is in one of those populations, namely $[p]$, which is exactly the population of all the people in the same country as $p$! $[p]$ might contain many people, or just $p$ alone, but it does exist. –  MJD Mar 21 '14 at 12:33
So there is no person in Europe who is not in one of these parts. And there is no person in Europe who is in more than one of these parts. So the parts exactly partition the population of Europe. –  MJD Mar 21 '14 at 12:39