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Lagrange thm: orbits as equivalence classes and cosets

buckeye1973

New member
Feb 22, 2012
5
Hi all, first post, please bear with me!

I am trying to understand Lagrange's Theorem by working through some exercises relating to the Orbit-Stabilizer Theorem (which I also do not fully understand.) I think essentially I'm needing to learn how to show cosets are equivalent to other things or each other.

Here's the setup:
Let $G$ be a group and $H$ be a subgroup of $G$. Define a relation on $G$ by:

$x \equiv_H y$ if and only if $x^{-1}y \in H$

I have already shown that $\equiv_H$ is an equivalence relation. Now I need to show that the $\equiv_H$ equivalence classes are the left cosets $gH$ of $H$ in $G$.

Here is my try:

Let $x,y \in G$ such that $x \equiv_H y$.
Then $x^{-1}y \in H$, which implies $(x^{-1}y)^{-1} = y^{-1}x \in H$.
So $y^{-1}x = h_y$ for some $h_y \in H$.
Then $y(y^{-1}x) = yh_y \in yH$.
thus $x = yh_y \in yH$.
Now $e \in H$, so there exists $h_x \in H$ such that $xh_x = yh_y$.

Um, Q.E.D.?

The logic flow doesn't feel right to me here. My other plan was to assume $xH \neq yH$ and show that that implies $x \not\equiv_H y$, but I couldn't even get that off the ground.

I'd really appreciate any ideas I can get here. Thanks!

Brian
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi all, first post, please bear with me!

I am trying to understand Lagrange's Theorem by working through some exercises relating to the Orbit-Stabilizer Theorem (which I also do not fully understand.) I think essentially I'm needing to learn how to show cosets are equivalent to other things or each other.

Here's the setup:
Let $G$ be a group and $H$ be a subgroup of $G$. Define a relation on $G$ by:

$x \equiv_H y$ if and only if $x^{-1}y \in H$

I have already shown that $\equiv_H$ is an equivalence relation. Now I need to show that the $\equiv_H$ equivalence classes are the left cosets $gH$ of $H$ in $G$.

Here is my try:

Let $x,y \in G$ such that $x \equiv_H y$.
Then $x^{-1}y \in H$, which implies $(x^{-1}y)^{-1} = y^{-1}x \in H$.
So $y^{-1}x = h_y$ for some $h_y \in H$.
Then $y(y^{-1}x) = yh_y \in yH$.
thus $x = yh_y \in yH$.

Since \(x\equiv_{\small{H}}y\Rightarrow x\in yH\) we can conclude that this equivalence relation divides the group into left cosets. In other words if x and y are equivalent to each other then both belong to the same coset, yH. You can finish the proof here.

Now $e \in H$, so there exists $h_x \in H$ such that $xh_x = yh_y$.


The statement written above(highlighted in red) is correct. But it does not directly imply that both x and y belong to the same coset of G.

Um, Q.E.D.?

The logic flow doesn't feel right to me here. My other plan was to assume $xH \neq yH$ and show that that implies $x \not\equiv_H y$, but I couldn't even get that off the ground.

I'd really appreciate any ideas I can get here. Thanks!

Brian
...
 

buckeye1973

New member
Feb 22, 2012
5
Aha, I see, then. The obvious fact that $y \in yH$ (by the identity in the subgroup $H$) had not come to mind. So my proof concludes:

Let $x,y \in G$ such that $x \equiv_H y$.
Then $x^{-1}y \in H$, which implies $(x^{-1}y)^{-1} = y^{-1}x \in H$.
So $y^{-1}x = h_y$ for some $h_y \in H$.
Then $y(y^{-1}x) = yh_y \in yH$.
thus $x = yh_y \in yH$.
Now, $y \in yH$ (because $e \in H$,) so $x$ and $y$ belong to the same left coset $yH$.
Further, by symmetry of the equivalence relation, $x$ and $y$ belong to the same left coset $xH$.
Therefore, $x \equiv_H y \Rightarrow xH = yH$.
In other words, the $\equiv_H$ equivalence classes are the left cosets $gH$ of $H$ in $G$.
$\square$


I think this is correct, yes?

Thanks again for your help!

Brian
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Aha, I see, then. The obvious fact that $y \in yH$ (by the identity in the subgroup $H$) had not come to mind. So my proof concludes:

Let $x,y \in G$ such that $x \equiv_H y$.
Then $x^{-1}y \in H$, which implies $(x^{-1}y)^{-1} = y^{-1}x \in H$.
So $y^{-1}x = h_y$ for some $h_y \in H$.
Then $y(y^{-1}x) = yh_y \in yH$.
thus $x = yh_y \in yH$.
Now, $y \in yH$ (because $e \in H$,) so $x$ and $y$ belong to the same left coset $yH$.
Further, by symmetry of the equivalence relation, $x$ and $y$ belong to the same left coset $xH$.
Therefore, $x \equiv_H y \Rightarrow xH = yH$.
In other words, the $\equiv_H$ equivalence classes are the left cosets $gH$ of $H$ in $G$.
$\square$


I think this is correct, yes?

Thanks again for your help!

Brian
Hi Brian,

Yes. I don't see any mistake in it.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
there's nothing wrong with what you did, but your proof implicitly relies on the fact that equivalence classes of an equivalence relation are disjoint (they partition the set they are defined on, the group G in this case).

i would be inclined to prove it like so: suppose x~y (where ~ is the given equivalence via H).

suppose xh is any element of xH. since x~y, $x^{-1}y \in H$, call it h'.

now H is a subgroup, so $h = eh = h'h'^{-1}h$, and $h'^{-1}h$ is another element of H, call it h".

then $xh = xh'h'' = x(x^{-1}y)h'' = yh''$, which is in yH, so xH is contained in yH.

using the fact that y~x (by symmetry of ~), we can prove yH is contained in xH in a similar fashion, thus xH = yH.

another minor point: it could perhaps be, that there are fewer left cosets than equivalence classes, that the partitioning of G into equivalence classes is a refinement of the partition into left cosets. so you need to show this, too:

if xH = yH, then x~y. this isn't too hard, we have that for every xh in xH, xh = yh', for some h' in H. multiplying both sides by $x^{-1}$ on the left:

$h = x^{-1}yh'$ and multiplying both sides by $h'^{-1}$ on the right, we have:

$hh'^{-1} = x^{-1}y$, and by closure (since H is a subgroup), the LHS is in H, so x~y.