KaKb=Kab
 

In grioup theory
If K is normal in G


KaKb=Kab


Which is the proof?

Step 1: prove that KK=K.

I am following a book
 

[QUOTE=Nick;517452]Step 1: prove that KK=K.[/QUOTE]



thge book says given Ka=Ka1 and Kb=Kb1 we must show that Kab=Ka1b1, equivalently that ab(a1b1)^(-1) belongs to K.
This last statement can be proven so:


if Ka=Ka1 and Kb=Kb1,


then Kab=Ka1b1, right cancelletion gives Kab(a1b1)^(-1)=K. So ab(a1b1)^(-1) must belong to K because of the property of closure of the subgroup K? or better Kab=Ka1b1 then by left cancellation ab(a1b1)^(-1)=1 belonging to K?

I don't undestand this equivalence
 

[QUOTE=enzocreti;517453]thge book says given Ka=Ka1 and Kb=Kb1 we must show that Kab=Ka1b1, equivalently that ab(a1b1)^(-1) belongs to K.
This last statement can be proven so:


if Ka=Ka1 and Kb=Kb1,










I dont understand this equivalence:


Kab=Ka1b1 is equivalent to say that ab(a1b1)^(-1) belongs to K

Your book appears to [B]define[/B] KaKb=Kab and then prove that this is well-defined, i.e. it does not depend on the choices of a and b.
That is not wrong but it is unnecessarily complicated.

It is simpler to define the product of A and B for any subsets A, B of the group G as follows:
\[AB=\{ab:a\in A, b\in B\}\]
and then prove as a result that KaKb=Kab if K is a normal subgroup of G.

how to proof
 

how to proof that the inverse of Ka when K is not normal is a^(-1)K?

If K is a subgroup of a group G then K is closed under the group operation and K contains the neutral/identity element of the group.
So for all \(a,b\in K\) we have \(ab\in K\) too, giving \(KK\subset K\)
and, for all \(a\in K\) we have \(a=a\cdot 1\in KK\) so \(K\subset KK\) as well
hence \(KK=K\).

Once you have that, everything is easy:
\(Kaa^{-1}K=K\cdot1\cdot K=KK=K. \)
And, if K is a normal subgroup of G then \(aK=Ka\) so
\( KaKb=KKab=Kab.\)