Why $G'' \triangleleft G$?

Question

Here is the question I want to solve:

Show that if $G'/G''$ and $G''/G'''$ are both cyclic then $G'' = G'''$. [you may assume $G''' = 1$. Then $G/G''$ acts by conjugation on the cyclic group $G''$.

Here is the solution I found here If $G'/G''$ and $G''/G'''$ are cyclic then $G''=G'''$ :

By the third isomorphism theorem, we may as well assume that $G''' = 1$. So we are given that $G'/G''$ and $G''/G''' = G''$ are cyclic, and we want to show that $G'' = 1$, or rather that $G'$ is abelian.

We will use the fact that for a subgroup $H$ of $G$, the quotient $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$.

Since $G''$ is normal, we have $G = N_G(G'')$. Then by the fact we just mentioned, $G/C_G(G'')$ is isomorphic to a subgroup of $\operatorname{Aut}(G'')$. But $G''$ is cyclic, so $\operatorname{Aut}(G'')$ is an abelian group. Thus $G/C_G(G'')$ is abelian, and so $G' \subset C_G(G'')$. But this means that $G'' \subset Z(G')$.

From here the proof follows from the fact

$$ N \subset Z(G), ~ G/N ~~ \text{cyclic} \Rightarrow G~~ \text{is abelian} ,$$

the proof of which is basically the same as the proof of the fact about $G/Z(G)$ you mentioned.

My question is:

Is there a rigor proof for the idea that $G'' \triangleright G$, in our case, that the author used in the answer? I do not know why this is correct, my professor even emphasized that $G^{(n)} \triangleleft G^{(n - 1)} \forall n$ but not necessarily $G^{(n)} \triangleleft G \forall n.$

Answer 1

Recall that if $K$ is a group, then $K'$ is the subgroup of $K$ generated by all elements of the form $[x,y]=x^{-1}y^{-1}xy$, as $x,y$ range over all elements of $K$. (If your convention as $[a,b]=aba^{-1}b^{-1}$, the argument is essentially the same.)

Definition. Let $H\leq G$. We say that $H$ is

1. Normal if and only if for every $g\in G$, $gHg^{-1}=H$. That is, for all $\phi\in\mathrm{Inn}(G)$, $\phi(H)=H$, where $\mathrm{Inn}(G)$ are the inner automorphisms of $G$.
2. Characterstic if and only if for every $\varphi\in\mathrm{Aut}(G)$, $\varphi(H)=H$.
3. Fully invariant if and only if for every $\phi\in\mathrm{End}(G)$, $\phi(H)\leq H$.

Note that fully invariant implies characteristic implies normal.

Proposition. Let $H\leq K\leq G$. If $K\triangleleft G$ and $H$ is characteristic in $K$, then $H\triangleleft G$.

Proof. Let $g\in G$. Then the map $\phi\colon K \to K$ given by $k\longmapsto gkg^{-1}$ is an automorphism of $K$, because $K$ is normal. Since $H$ is characteristic in $K$, it follows that $\phi(H)=H$. That is, for each $h\in H$, $ghg^{-1}\in H$.

Thus, for each $g\in G$, $h\in H$, $ghg^{-1}\in H$. Thus, $H\triangleleft G$, as claimed. $\Box$

Propostion. Let $K$ be a group. Then $K'$ is fully invariant in $K$; in particular, it is characteristic in $K$.

Proof. Since $K'$ is generated by elements $[x,y]$, it suffices to show that if $\phi\in\mathrm{End}(K)$, then for all $x,y\in K$ we have $\phi([x,y])\in K'$.

Indeed, $$\begin{align*} \phi([x,y]) &= \phi(x^{-1}y^{-1}xy)\\ &= \phi(x)^{-1}\phi(y)^{-1}\phi(x)\phi(y)\\ &= [\phi(x),\phi(y)] \in K'. \end{align*}$$ Thus, $\phi(K')\leq K'$, as claimed. $\Box$

Corollary. For every $n\geq 0$, $G^{(n)}\triangleleft G$.