# The group $G$ is a partition of equivalence classes

I am looking at the proof of the Lagrange Theorem:

Let $G$ a finite group and $H$ a subgroup of G. Then $|H| \mid |G|$.

If $g \in G$, then $|gH|=|H|$.

If $g_1, g_2 \in G$ we consider $g_1H$ and $g_2H$.

To continue we show that $$g_1H \cap g_2H=\varnothing \text{ or } g_1H=g_2H$$ as followed:

At the set $G$ we define a relation $\sim$ like that: $$g_1 \sim g_2 \Leftrightarrow g_2^{-1} g_1 \in H$$ Then we show that this relation $\sim$ is an equivalence relation.

So $G$ is a partition of equivalence classes.

(partition=union of disjoint subsets)

 Could explain me why when we know that there is an equivalence relation then the group $G$ is a partition of equivalence classes?

This has nothing to do with the algebraic property of a group, in general if one can define an equivalence relation on a set, then the set partitions into equivalence classes. Otherwise put, the set is the disjoint union of its equivalence classes. For that matter, $G$ could also have been a topological space for example. In your case the group structure itself gives rise to an equivalence relation.

• But...how did we find this equivalence relation? – evinda May 26 '14 at 10:50
• That is an excellent question, hard to answer. There is no generic way of constructing equivalence relations. In fact, on a set of $n \geq 0$ elements, the number of equivalence relations equals $B_n$, the $n$-th Bell number ($1, 1, 2, 5, 15, 52, 203, 877,...$), see en.wikipedia.org/wiki/Bell_numbers. In case of the $G$ and $H$ above, one could say that the equivalence relation deems elements at "the same distance" (measured by the subgroup $H$) equivalent, so a kind of abstraction notion of parallel lines. See also en.wikipedia.org/wiki/Equivalence_relation. Hope this helps. – Nicky Hekster May 26 '14 at 11:48

Suppose the intersection of $g_{1}H$ and $g_{2}H$ is non-empty. Then there are $h_{1}$ and $h_{2}$ s.t. $g_{1}h_{1}=g_{2}h_{2}$.

If $x\in g_{1}H$, then

$x=g_{1}h=g_{2}h_{2}h_{1}^{-1}h\in g_{2}H$.

Thus $g_{1}H$ is a subset of $g_{2}H$.

If $x\in g_{2}H$, then

$x=g_{2}h=g_{1}h_{1}h_{2}^{-1}h\in g_{1}H$.

Thus $g_{2}H$ is a subset of $g_{1}H$.

Hence

$g_{1}H=g_{2}H$

Let $g\in G$. Then

$g=ge\in gH$

This proves that the left cosets of H in G are disjoint and that G is the union of these cosets.

Together with a proof of $|gH|=|H|$ you are more or less done.

• So, this is an other way instead of using the equivalence relation, right?  So we show that if $g_1H$ and $g_2H$ have a common point, it should be $g_1H=g_2H$, else it should be $g_1H \cap g_2H =\varnothing$, or not?  And knowing that it is $g_1H=g_2H$ or $g_1H \cap g_2H = \varnothing$, do we conclude that $G$ is a partition of these cosets? Is it because this stands $\forall g \in G$ ? – evinda May 26 '14 at 10:54
• Yes, yes and yes. – poolpt May 26 '14 at 13:08