# Abstract Algebra (commutative property)

• Apr 19th 2013, 12:25 PM
mohammedlabeeb
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
• Apr 19th 2013, 12:40 PM
emakarov
Re: Abstract Algebra (commutative property)
Clearly, hk = kh iff hkh-1k-1 = e. Prove this by showing that hkh-1k-1 ∈ H ∩ K.
• Apr 20th 2013, 05:42 AM
mohammedlabeeb
Re: Abstract Algebra (commutative property)
Please I want would like to see more steps/details
• Apr 20th 2013, 07:50 AM
emakarov
Re: Abstract Algebra (commutative property)
Quote:

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.
• Apr 20th 2013, 08:22 AM
mohammedlabeeb
Re: Abstract Algebra (commutative property)
How do you know that $\displaystyle hkh^{-1}k^{-1} \in\ H intersection K$ ?
• Apr 20th 2013, 08:27 AM
Plato
Re: Abstract Algebra (commutative property)
Quote:

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

Do you know that $\displaystyle hkh^{-1}\in K~?$ HOW?

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

Now do as you have been told: do the same wrt $\displaystyle H$.
• Apr 20th 2013, 01:32 PM
mohammedlabeeb
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
Attachment 28052
Quote:

Originally Posted by Plato
Do you know that $\displaystyle hkh^{-1}\in K~?$ HOW?

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

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

• Apr 20th 2013, 01:39 PM
emakarov
Re: Abstract Algebra (commutative property)
Quote:

Originally Posted by Plato
Do you know that $\displaystyle hkh^{-1}\in K~?$ HOW?

Quote:

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?
• Apr 20th 2013, 02:37 PM
emakarov
Re: Abstract Algebra (commutative property)
Quote:

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.