Subgroups

[Nick]

For all \(x,y\in G\),
\[ (xy)(y^{-1}x^{-1})=x(yy^{-1})x^{-1}=x1x^{-1}=xx^{-1}=1=(xy)(xy)^{-1}\]
so (cancelling \((xy)\) on the left) we have \((xy)^{-1}=y^{-1}x^{-1}\).

This is sometimes called the "socks and shoes" rule: the opposite of putting on your socks then your shoes is to take off your shoes then your socks.

[enzocreti]

Originally Posted by Nick

For all \(x,y\in G\),
\[ (xy)(y^{-1}x^{-1})=x(yy^{-1})x^{-1}=x1x^{-1}=xx^{-1}=1=(xy)(xy)^{-1}\]
so (cancelling \((xy)\) on the left) we have \((xy)^{-1}=y^{-1}x^{-1}\).

This is sometimes called the "socks and shoes" rule: the opposite of putting on your socks then your shoes is to take off your shoes then your socks.

From this follows that


there is a bijection between Ha and a^(-1)H


Is this proof correct:


H=a^(-1)b^(-1)H
Hb=a^(-1)H
because Hb=Ha then
Ha=a^(-1)H


???

[Nick]

If \(Ha=Hb\) then \(H=ab^{-1}H\) not \(a^{-1}b^{-1}H\).

[enzocreti]

Originally Posted by Nick

If \(Ha=Hb\) then \(H=ab^{-1}H\) not \(a^{-1}b^{-1}H\).

Now I unerstand




if x=ha belongs to Ha so x^(-1) belongs to Ha


this implies x^(-1)=a^(-1)h^(-1)...so both x and x^(-1) belong to a^(-1)H...to complete the proof it is enough to proof the viceversa