Derived subgroup is normal Suppose we have a group $G$ with commutator subgroup $G'$. I have shown that $G'$ is a normal subgroup of $G$, but I want to show that $G^n$ is also a normal subgroup of $G$. Here we've defined $G^n$ recursively, i.e. $G^n = (G^{n-1})'$.
What I've done so far is: Let $g\in G$ and $x\in G'$. Then by definition $gxg^{-1}x^{-1} \in G'$, and furthermore since $x\in G'$, it follows that $gxg^{-1} = gxg^{-1}x^{-1}x \in G'$, hence $G'$ is normal in $G$.
I'm stuck where to go from here in proving that $G^n$ is also normal. Any help?
 A: Let $G''$ be the derived group of $G'$.  By definition, $G''$ is the subgroup of $G$ generated by elements of the form $z= xyx^{-1}y^{-1}$, where $x$ and $y$ are in $G'$.  
First try to show that $G''$ is normal in $G$.  This isn't too hard to do.  It suffices to show that if $z$ is one of these generators of $G''$, and $g \in G$, then $gzg^{-1}$ is also one of these generators.
A: Here’s a fun way of doing it. Use the following facts

  
*
  
*Any characteristic subgroup of a normal subgroup is normal in the whole group.
  
*The derived subgroup is characteristic.
  

then proceed by induction.
A: Your argument shows that $G^n$ is normal in $G^{n-1}$. We would like to argue that this inductively shows that $G^n$ is normal in $G$. Unfortunately, normality is not transitive, in the sense that if $K$ is normal in $H$ and $H$ is normal in $G$, $K$ may not be normal in $G$.
You should know some stronger property than normality that does have this transitivity. Perhaps you could show that $G'$ has this property to make the induction work.
