if xH=yH then H(x^(-1))=H(y^(-1))
 

Ho to proof this:


if xH=yH then H(x^(-1))=H(y^(-1))?

It depends where you want to start.
Are you OK with the following facts?
\(xH=yH\Leftrightarrow y^{-1}x\in H\)
\(Hu=Hv\Leftrightarrow uv^{-1}\in H\)

...
 

[QUOTE=Nick;525630]It depends where you want to start.
Are you OK with the following facts?
\(xH=yH\Leftrightarrow y^{-1}x\in H\)
\(Hu=Hv\Leftrightarrow uv^{-1}\in H\)[/QUOTE]

yes then?

If xH=yH then \(y^{-1}x\in H\)
and H is a subgroup so the inverse of \(y^{-1}x\) is also an element of H, i.e. \(x^{-1}y\in H\).
Let \(u=x^{-1}\) and \(v=y^{-1}\).
Then \(v^{-1}=y\) so \(uv^{-1}\in H\) and therefore \(Hu=Hv\) i.e. \(Hx^{-1}=Hy^{-1}\).

[QUOTE=Nick;525644]If xH=yH then \(y^{-1}x\in H\)
and H is a subgroup so the inverse of \(y^{-1}x\) is also an element of H, i.e. \(x^{-1}y\in H\).
Let \(u=x^{-1}\) and \(v=y^{-1}\).
Then \(v^{-1}=y\) so \(uv^{-1}\in H\) and therefore \(Hu=Hv\) i.e. \(Hx^{-1}=Hy^{-1}\).[/QUOTE]
ok thanks


so the mapping left cosets right cosets is a bijection...

[QUOTE=enzocreti;525646]so the mapping left cosets right cosets is a bijection...[/QUOTE]
Yes, so you can define the index of H in G, written [G:H], as the number of left cosets or the number of right cosets of H in G.

the number of left cosets is equal to the number of the right cosets
 

[QUOTE=Nick;525679]Yes, so you can define the index of H in G, written [G:H], as the number of left cosets or the number of right cosets of H in G.[/QUOTE]




So the number of left cosets is always equal to the number of right cosets, because of the bijection?

[QUOTE=enzocreti;525718]So the number of left cosets is always equal to the number of right cosets, because of the bijection?[/QUOTE]
Yes, if a bijection exists between 2 sets then it follows that they have the same number of elements.