1. ## Commutator, Abelian

A commutator in a group is any element $[x, y]=xyx^{-1}y^{-1}$ for $x, y \in G$. If $K$ is subgroup of $G$, prove $K$ contains all the commutators of $G$ iff $K \triangleleft G$ and $G/K$ is abelian.
Attempt
( $\Rightarrow$) $K \triangleleft G$ is trivial. To show $G/K$ is abelian, take $g_1 +K$ and $g_2+K$ and show they commute, how do I do this?
( $\Leftarrow$) This is the harder direction. I don't see how to prove this way.

Thanks for any help.

2. Originally Posted by canberra1454
A commutator in a group is any element $[x, y]=xyx^{-1}y^{-1}$ for $x, y \in G$. If $K$ is subgroup of $G$, prove $K$ contains all the commutators of $G$ iff $K \triangleleft G$ and $G/K$ is abelian.
Attempt
( $\Rightarrow$) $K \triangleleft G$ is trivial. To show $G/K$ is abelian, take $g_1 +K$ and $g_2+K$ and show they commute, how do I do this?
( $\Leftarrow$) This is the harder direction. I don't see how to prove this way.

Thanks for any help.
( $\Rightarrow$) If $K$ contains all the commutators of $G$, then $K \triangleleft G$ and $G/K$ is abelian.

Assume K contains all the commutators of $G$. Then $xyx^{-1}y^{-1} \in K$ for all $x,y \in G$. It follows that $(xy)(yx)^{-1} \in K$. Thus, $(xy)K = (yx)K$ for all $x,y \in G$. Therefore, $G/K$ is abelian.
We remain to show that K is normal. To show that K is normal, we need to show that for all $g \in G$, $gKg^{-1} = K$. Let $x^{-1}y^{-1}xy \in K$ for $x,y \in G$. Then, we need to check $g(x^{-1}y^{-1}xy)g^{-1}$ is in K. $g(x^{-1}y^{-1}xy)g^{-1} = gx^{-1}y^{-1}x(g^{-1}yy^{-1}g)yg^{-1}=((gx^{-1})y^{-1}(gx^{-1})^{-1}y)(y^{-1}gyg^{-1})$, which is in K. Thus, $K \triangleleft G$.

( $\Leftarrow$) If $K \triangleleft G$ and $G/K$ is abelian, then $K$ contains all the commutators of $G$.

Assume $K \triangleleft G$ and $G/K$ is abelian. Then, $(xy)K = (yx)K$ for all $x,y \in G$. It follows that $(xy)(yx)^{-1} = xyx^{-1}y^{-1} \in K$. Thus, K contains all the commutator of G.

3. Originally Posted by aliceinwonderland
We remain to show that K is normal. To show that K is normal, we need to show that for all $g \in G$, $gKg^{-1} = K$. Let $x^{-1}y^{-1}xy \in K$ for $x,y \in G$. Then, we need to check $g(x^{-1}y^{-1}xy)g^{-1}$ is in K. $g(x^{-1}y^{-1}xy)g^{-1} = gx^{-1}y^{-1}x(g^{-1}yy^{-1}g)yg^{-1}=((gx^{-1})y^{-1}(gx^{-1})^{-1}y)(y^{-1}gyg^{-1})$, which is in K. Thus, $K \triangleleft G$.
I was working on this problem and noticed a slight error, but an easy fix. You assumed that the element you were testing the conjugation by g with was in the commutator subgroup (G'), but we only know that K contains the commutator subgroup. So the correct proof should go something like this.

Let $k\in K$ and for any $g\in G$ we consider $gkg^{-1}$ and show it is in K.

$gkg^{-1}k^{-1}\in G' \subset K$

As K is a subgroup, it is closed under multiplication, so
$gkg^{-1}k^{-1}k = gk^{-1}g^{-1} \in K$.

g and k were arbitraty, so $gKg^{-1} \subset K \Rightarrow K \triangleleft G$ as desired.

4. Originally Posted by Gamma
I was working on this problem and noticed a slight error, but an easy fix. You assumed that the element you were testing the conjugation by g with was in the commutator subgroup (G'), but we only know that K contains the commutator subgroup. So the correct proof should go something like this.

Let $k\in K$ and for any $g\in G$ we consider $gkg^{-1}$ and show it is in K.

$gkg^{-1}k^{-1}\in G' \subset K$

As K is a subgroup, it is closed under multiplication, so
$gkg^{-1}k^{-1}k = gk^{-1}g^{-1} \in K$.

g and k were arbitraty, so $gKg^{-1} \subset K \Rightarrow K \triangleleft G$ as desired.
hmm, since K contains all the commutators of G, isn't the below one in my previous post an enough argument to show that K is normal in G?

"To show that K is normal, we need to show that for all $g \in G$, $gKg^{-1} = K$. Let $x^{-1}y^{-1}xy \in K$ for $x,y \in G$. Then, we need to check $g(x^{-1}y^{-1}xy)g^{-1}$ is in K."

I think the only step I missed is to show that K is indeed a subgroup of G, but it is trivial.

5. What if $K-G'\not = \emptyset$? Not every element in K needs to be commutator is what my concern is.

6. Originally Posted by Gamma
What if $K-G'\not = \emptyset$? Not every element in K needs to be commutator is what my concern is.
aha, I read the question again and I found that K is a subgroup of G by hypothesis. So, K can be a subgroup of G that contains G'.

Thanks for pointing out my mistake.

7. Bingo! I was trying to do the exact same thing you did when I was going through example problems, but thought something was fishy since that is basically exactly how I proved G' was normal in G, so I thought something was up. It is kind of a cute proof IMHO.

Seems like a fairly common algebra question, so just wanted to clean it up in case someone like myself was searching for inspiration for it online.