Herstein Question: $G^{i}$ normal in $G$? I just wanted to ask a quick question. I'm going over the second edition of I.N. Herstein's topics in algebra and one of his exercises asks the reader to prove that each $G^{i} $ is a normal subgroup of G where $G^{i}$ is the $i^{th}$ commutator group.
Now I think I was able to prove $G^{i}$ is normal in $G^{i+1}$, where $G^{i}$ is the commutator group of $G^{i+1}$:
Let $g \in G^{i+1}, h \in G^{i}$. Every $h \in G^{i+1}$ since $h=g_{1}^{-1}g_{2}^{-1}g_{1}g_{2} \in G^{i+1}$ since $G^{i+1}$ is a group.
So
$(gh_{1}g^{-1}h_{1}^{-1} \in G^{i}) \Rightarrow (gh_{1}g^{-1}h_{1}^{-1}=h_{2}) \Rightarrow (gh_{1}=h_{2}h_{1}g) \Rightarrow (gh_{1}=h_{3}g)$
So $G^{i}$ is normal in $G^{i+1}$.
I'm seeing how this relates to solvable groups through composition series. However it doesn't seem necessary to have every $G^{i}$ normal to the whole group G. We only really need $G^{i}$ normal in $G^{i+1}$. Is it absolutely necessary I prove the additional property of its normality in G? I suppose for the sake of curiosity I might come back to prove the proposition but, for now, can I move on without being delayed by it? I'd like to get to the more exciting stuff. Also is my proof correct?
Thanks in advance!!
 A: Well, you surely know that the derived subgroup $G'$ is normal, actually characteristic, in $G$. So $G^{i+1} = (G^{i})'$ is characteristic in $G^{i}$.
Now, proceeding by induction, assuming $G^{i}$ normal in $G$, since $G^{i+1}$ is characteristic in $G^{i}$, we have that $G^{i+1}$ is normal in $G$.

Addendum Perhaps it could be mentioned that each $G^{i}$ is a fully invariant subgroup of $G$, that is, it is sent into itself by any endomorphism of $G$. The reason is that $G^{i}$ is a verbal subgroup, i.e., it is generated by the values taken in $G$ by a certain word. The word in question is
$$
[x_{1}, x_{2}]
$$
for $G^{2}$,
$$
[[x_{1}, x_{2}],[x_{3},x_{4}]]
$$
for $G^{3}$, and so on. Here $[y,z] = y^{-1} z^{-1} y z$ is the commutator.
A: The claim is still true because the commutator subgroup is even more special than a normal subgroup: it is also a characteristic subgroup.
Then you can easily show another lemma: 

A characteristic subgroup of a characteristic subgroup of $G$ is a characteristic subgroup of $G$.

Then you will have shown that all the derived subgroups of G are characteristic, hence normal in G.

There is another little exercise related to the above: a characteristic subgroup of a normal subgroup of G is a normal subgroup of G.
A: Hint: Prove $G^{i+1}$ is characteristic in $G^i$. This provides an induction step.
A: Since $[H,K]^g = [H^g, K^g]$, it follows that
$$(G^i)^g = [(G^{i-1})^g, (G^{i-1})^g]$$
so clearly $G^i$ is a normal subgroup.
