# Math Help - Abstract Algebra (commutative property)

1. ## Abstract 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

2. ## Re: Abstract Algebra (commutative property)

Clearly, hk = kh iff hkh-1k-1 = e. Prove this by showing that hkh-1k-1 ∈ H ∩ K.

3. ## Re: Abstract Algebra (commutative property)

Please I want would like to see more steps/details

4. ## Re: Abstract Algebra (commutative property)

Originally Posted by emakarov
Clearly, hk = kh iff hkh-1k-1 = e. Prove this by showing that hkh-1k-1 ∈ H ∩ K.
Do you agree that showing hkh-1k-1 ∈ H ∩ K is sufficient? To prove that hkh-1k-1 ∈ H, break hkh-1k-1 into two factors both of which are in H, and do similarly for K.

5. ## Re: Abstract Algebra (commutative property)

How do you know that $hkh^{-1}k^{-1} \in\ H intersection K$ ?

6. ## Re: Abstract Algebra (commutative property)

Originally Posted by mohammedlabeeb
How do you know that $hkh^{-1}k^{-1} \in\ H intersection K$ ?
Do you know that $hkh^{-1}\in K~?$ HOW?

Then, do you know that $\left[hkh^{-1}\right]k^{-1}\in K~?$ HOW?.

Now do as you have been told: do the same wrt $H$.

7. ## Re: 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
CCI04202013_00003.bmp
Originally Posted by Plato
Do you know that $hkh^{-1}\in K~?$ HOW?

Then, do you know that $\left[hkh^{-1}\right]k^{-1}\in K~?$ HOW?.

Now do as you have been told: do the same wrt $H$.

8. ## Re: Abstract Algebra (commutative property)

Originally Posted by Plato
Do you know that $hkh^{-1}\in K~?$ HOW?
Originally Posted by mohammedlabeeb
I don't know How?!. Can you tell me How?
In another thread we have already discussed the definition of a normal subgroup. Don't you recognize it in Plato's quote?

9. ## Re: Abstract Algebra (commutative property)

Originally Posted by mohammedlabeeb
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
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 h2k2 = k2h2. You assumed what you are supposed to prove.

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