Subgroups
 

Let be H a subgroup of a Group G.
If Ha=Hb it follows that a^(-1)H=b^(-1)H


Because ha=h'b, it follows that b=h^(-1)h'a.
So b belongs to Ha.
and b^(-1)=a^(-1)hh'^(-1). So b^(-1) belongs to a^(-1)H.
So there is a bijection between b^(-1) and a^(-1)H and this proves that if Ha=Hb then a^(-1)H=b^(-1)H????
And what about the bijection Ha>a^(-1)H?

[QUOTE=enzocreti;516732]Let be H a subgroup of a Group G.
If Ha=Hb it follows that a^(-1)H=b^(-1)H
[/QUOTE]
Hint: \(H^{-1}=H\).

And so?
 

[QUOTE=Nick;516734]Hint: \(H^{-1}=H\).[/QUOTE]




And so?

[QUOTE=enzocreti;516735]And so?[/QUOTE]
For more detail, see the proof of proposition 82 in our Number Theory discussion subforum[URL="https://www.mersenneforum.org/showthread.php?t=21904"] here[/URL]

subgroup
 

[QUOTE=Nick;516736]For more detail, see the proof of proposition 82 in our Number Theory discussion subforum[URL="https://www.mersenneforum.org/showthread.php?t=21904"] here[/URL][/QUOTE]


H=Hab^(-1) only if Ha=Hb?

[QUOTE=enzocreti;516753]H=Hab^(-1) only if Ha=Hb?[/QUOTE]
Yes, if and only if.

now it is clearer
 

yes now it is clearer thank you


Ha=Hb


Hab^(-1)=H

If xH=yH, then Hx^(-1)=Hy^(-1).
 

I tried a proof:


If xH=yH then xh=yh' for some h and h' belonging to H.


This yields:
h'h^(-1)=y^(-1)(x^(-1))^(-1)


...follows h^(-1)x^(-1)=h'y^(-1)


because this must be true for all h and h' belonging to H, then Hx^(-1)=Hy^(-1)


Is this correct?

[QUOTE=enzocreti;516815]Is this correct?[/QUOTE]
Nearly!

Suppose \(xH=yH\).
Take any \(z\in Hx^{-1}\). Then \(z=h_1x^{-1}\) for some \(h_1\in H\).
So \(z^{-1}=(h_1x^{-1})^{-1}=xh_1^{-1}\) and \(h_1^{-1}\in H\) so \(z^{-1}\in xH\).
But \(xH=yH\) so \(z^{-1}=yh_2\) for some \(h_2\in H\) as well.
Hence \(z=(yh_2)^{-1}=h_2^{-1}y^{-1}\) and \(h_2^{-1}\in H\) so \(z\in Hy^{-1}\).
This proves that \(Hx^{-1}\subset Hy^{=1}\), and \(Hy^{-1}\subset Hx^{-1}\) follows by symmetry, so they are equal.

!
 

Nice proof now I completely understood! Thanks

only one thing...
 

only one thing I don't understand well


3rd row:


z^(-1)=(h1x^(-1))^(-1)=xh1^(-1)


it couldnt be the inverse? = h1^(-1)x? so z^(-1) couldnt belong to Hx instead of xH?