1. ## Commutator, Abelian

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

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

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

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

Assume $\displaystyle K \triangleleft G$ and $\displaystyle G/K$ is abelian. Then, $\displaystyle (xy)K = (yx)K$ for all $\displaystyle x,y \in G$. It follows that $\displaystyle (xy)(yx)^{-1} = xyx^{-1}y^{-1} \in K$. 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 $\displaystyle g \in G$,$\displaystyle gKg^{-1} = K$. Let $\displaystyle x^{-1}y^{-1}xy \in K$ for $\displaystyle x,y \in G$. Then, we need to check $\displaystyle g(x^{-1}y^{-1}xy)g^{-1}$ is in K. $\displaystyle g(x^{-1}y^{-1}xy)g^{-1} = gx^{-1}y^{-1}x(g^{-1}yy^{-1}g)yg^{-1}=((gx^{-1})y^{-1}(gx^{-1})^{-1}y)(y^{-1}gyg^{-1})$, which is in K. Thus, $\displaystyle K \triangleleft G$.
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 $\displaystyle k\in K$ and for any $\displaystyle g\in G$ we consider $\displaystyle gkg^{-1}$ and show it is in K.

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

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

g and k were arbitraty, so $\displaystyle gKg^{-1} \subset K \Rightarrow K \triangleleft G$ 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 $\displaystyle k\in K$ and for any $\displaystyle g\in G$ we consider $\displaystyle gkg^{-1}$ and show it is in K.

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

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

g and k were arbitraty, so $\displaystyle gKg^{-1} \subset K \Rightarrow K \triangleleft G$ 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 $\displaystyle g \in G$,$\displaystyle gKg^{-1} = K$. Let $\displaystyle x^{-1}y^{-1}xy \in K$ for $\displaystyle x,y \in G$. Then, we need to check $\displaystyle g(x^{-1}y^{-1}xy)g^{-1}$ 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 $\displaystyle K-G'\not = \emptyset$? Not every element in K needs to be commutator is what my concern is.

6. Originally Posted by Gamma
What if $\displaystyle K-G'\not = \emptyset$? 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.