Suppose H and K are Normal Subgroups of G and ( H intersection K is the identity). Prove that hk=kh for all h in H and k in K

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- Apr 19th 2013, 12:25 PMmohammedlabeebAbstract Algebra (commutative property)
Suppose H and K are Normal Subgroups of G and ( H intersection K is the identity). Prove that hk=kh for all h in H and k in K

- Apr 19th 2013, 12:40 PMemakarovRe: Abstract Algebra (commutative property)
Clearly, hk = kh iff hkh

^{-1}k^{-1}= e. Prove this by showing that hkh^{-1}k^{-1}∈ H ∩ K. - Apr 20th 2013, 05:42 AMmohammedlabeebRe: Abstract Algebra (commutative property)
Please I want would like to see more steps/details

- Apr 20th 2013, 07:50 AMemakarovRe: Abstract Algebra (commutative property)
- Apr 20th 2013, 08:22 AMmohammedlabeebRe: Abstract Algebra (commutative property)
How do you know that $\displaystyle hkh^{-1}k^{-1} \in\ H intersection K$ ?

- Apr 20th 2013, 08:27 AMPlatoRe: Abstract Algebra (commutative property)
- Apr 20th 2013, 01:32 PMmohammedlabeebRe: Abstract Algebra (commutative property)
I don't know How?!. Can you tell me How?

I attached my solution I think it is not correct but you might be able to see what I did and tell me how to correct it

Attachment 28052 - Apr 20th 2013, 01:39 PMemakarovRe: Abstract Algebra (commutative property)
In another thread we have already discussed the definition of a normal subgroup. Don't you recognize it in Plato's quote?

- Apr 20th 2013, 02:37 PMemakarovRe: Abstract Algebra (commutative property)
I don't think it works. First, you did not properly introduce a and b: what exactly are they? More seriously, you started with hk = kh and arrived at h

_{2}k_{2}= k_{2}h_{2}. You assumed what you are supposed to prove.

The idea is to note that hk = kh ⇔ hkh^{-1}k^{-1}= e ⇔ (hkh^{-1}k^{-1}∈ H and hkh^{-1}k^{-1}∈ K). This has been described several times in this thread.