# Group theory question

1. Jan 21, 2008

### ehrenfest

[SOLVED] group theory question

1. The problem statement, all variables and given/known data
Prove that the commutator subgroup is normal.

2. Relevant equations

3. 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.

2. Jan 21, 2008

### Mystic998

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

3. Jan 21, 2008

### 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:

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

4. Jan 21, 2008

### 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.

5. Jan 21, 2008

### ehrenfest

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.

6. Jan 21, 2008

### 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.

7. Jan 21, 2008

### Mathdope

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.