Subgroups

[enzocreti]

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?

[Nick]

Originally Posted by enzocreti

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

Hint: \(H^{-1}=H\).

[enzocreti]

Originally Posted by Nick

Hint: \(H^{-1}=H\).

And so?

[Nick]

Originally Posted by enzocreti

And so?

For more detail, see the proof of proposition 82 in our Number Theory discussion subforum here

[enzocreti]

Originally Posted by Nick

For more detail, see the proof of proposition 82 in our Number Theory discussion subforum here

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

[Nick]

Originally Posted by enzocreti

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

Yes, if and only if.

[enzocreti]

yes now it is clearer thank you


Ha=Hb


Hab^(-1)=H

[enzocreti]

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?

[Nick]

Originally Posted by enzocreti

Is this correct?

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.

[enzocreti]

Nice proof now I completely understood! Thanks

[enzocreti]

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?