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.

  • 1
    $\begingroup$ But...how did we find this equivalence relation? $\endgroup$ – evinda May 26 '14 at 10:50
  • $\begingroup$ 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. $\endgroup$ – 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 $g_{1}H$.



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.

  • 1
    $\begingroup$ 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 $\forall g \in G$ ? $\endgroup$ – evinda May 26 '14 at 10:54
  • 1
    $\begingroup$ Yes, yes and yes. $\endgroup$ – poolpt May 26 '14 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.