KaKb=Kab

enzocreti · 2019-05-22, 06:07

In grioup theory
If K is normal in G

KaKb=Kab

Which is the proof?

Nick · 2019-05-22, 07:39

Step 1: prove that KK=K.

enzocreti · 2019-05-22, 07:44

Quote:

Originally Posted by Nick

Step 1: prove that KK=K.

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?

enzocreti · 2019-05-22, 07:56

[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

Nick · 2019-05-22, 13:15

Your book appears to define 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.

enzocreti · 2019-05-22, 13:32

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

Nick · 2019-05-22, 14:32

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
\( KaKb=KKab=Kab.\)

2019-05-22, 06:07	#1
enzocreti Mar 2018 1000010010₂ Posts	KaKb=Kab In grioup theory If K is normal in G KaKb=Kab Which is the proof?

2019-05-22, 07:56	#4
enzocreti Mar 2018 2×5×53 Posts	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

2019-05-22, 13:32	#6
enzocreti Mar 2018 2·5·53 Posts	how to proof how to proof that the inverse of Ka when K is not normal is a^(-1)K?

2019-05-22, 07:39	#2
Nick Dec 2012 The Netherlands 2·23·37 Posts	Step 1: prove that KK=K.

2019-05-22, 13:15	#5
Nick Dec 2012 The Netherlands 2×23×37 Posts	Your book appears to define 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.

2019-05-22, 14:32	#7
Nick Dec 2012 The Netherlands 2×23×37 Posts	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 \( KaKb=KKab=Kab.\)