Proving Normality of the Commutator Subgroup in Group Theory [SOLVED]

[ehrenfest]

[SOLVED] group theory question

Homework Statement

Prove that the commutator subgroup is normal.

Homework Equations

The Attempt at a Solution

Let H be the subgroup generated by all of the commutators. We want to show that H is normal.
Let x be in yH. Then, $y^{-1}x=aba^{-1}b^{-1}$ for some a,b in G.
So, $xy^{-1}=xaba^{-1}b^{-1} x^{-1}$
I am not seeing how to transform that into a commutator.

 

[Mystic998]

Try interpolating some $xx^{-1}$'s or $x^{-1}x$'s in the multiplication.

 

[ehrenfest]

I actually made a mistake in the original post. I need to express $xaba^{-1}b^{-1} x^{-1}$ as a product of powers of commutators. I BELIEVE THAT IT IS NOT IN GENERAL POSSIBLE TO EXPRESS THAT AS A SINGLE COMMUTATOR.
What you need to do is this:

$$xaba^{-1}b^{-1} x^{-1} = xaba^{-1}(x^{-1}b^{-1}bx) b^{-1} x^{-1} = (xa)b(xa)^{-1}b^{-1}(bxb^{-1}x^{-1})$$

 

[Mathdope]

That looks right but I wonder why the problem said the product of POWERS of commutators?

Since you've shown that it's expressible as a product of commutators, you have shown that it's in the group generated by the commutators, and hence that group is normal. Well done.

 

[ehrenfest]

That looks right but I wonder why the problem said the product of POWERS of commutators?

The problem did not say that. The problem only said show that the commutator subgroup is normal.

The definition of the commutator subgroup, however, is the collection of all products of powers of commutators.

 

[Mystic998]

I thought that the product of 2 commutators was a commutator itself. Of course, I haven't actually bothered to verify that, so I could be completely wrong.

 

[Mathdope]

I actually made a mistake in the original post. I need to express $xaba^{-1}b^{-1} x^{-1}$ as a product of powers of commutators.

The problem did not say that. The problem only said show that the commutator subgroup is normal.

The definition of the commutator subgroup, however, is the collection of all products of powers of commutators.

Got it. So you've shown that it's in the group generated by the commutators, but the top statement is where you mentioned powers. Obviously, from your answer, you got it without using powers.