We say a tower $G=G_0\supseteq G_1 \supseteq G_2\dots \supseteq G_m$ is abelian if it is normal (i.e., each $G_{i+1}$ is normal in $G_i$) and if each factor group $G_i/G_{i+1}$ is abelian.
I am working on the following exercise.
Let $G$ be a solvable group. (That means group $G$ has an abelian tower, whose last element is the trivial subsgroup.) Show that there exists an abelian tower $G=G_0\supseteq G_1 \supseteq G_2\dots \supseteq G_m=\{e\}$ so that each $G_i$ is normal subgroup of $G$?
I just try to verify the definition of normal subgroup. For every $g\in G$, I try to show that
$$
gG_{i}g^{-1}=G_i
$$
Since each $G_{i}$ is normal subgroup in $G_{i-1}$, then for every $g_{i-1}\in G_{i-1}$, we have
$$
g_{i-1}G_{i}g_{i-1}^{-1}=G_{i}
$$
Then repeat this one we get for $g_{i-2}\in G_{i-2}$, $$g_{i-2}G_{i-1}g_{i-1}^{-1}=G_{i-1}$$.
I am stuck here. But I did not use the condition 'each factor group $G_i\setminus G_{i+1}$ is abelian'…
I also find this note:https://users.math.msu.edu/users/ruiterj2/Math/Documents/Fall%202016/Algebra/Theorems.pdf
In page 9,
Theorem 1.81 Every $p$-group is solvable.
Theorem 1.82 Let $G$ be a non-trivial $p$-group. Then there exists a sequence of subgroups
$$G=G_0\supseteq G_1 \supseteq G_2\dots \supseteq G_m=\{e\}$$
such that $G_i$ is normal in $G$ and $G_i/G_{i+1}$ is cyclic of order $p$.
Best Answer
You seem to be trying to prove that if $G$ is solvable, then in any abelian tower all subgroups that appear in the tower are normal in the whole group. That is not true. What is true is that if $G$ is solvable, then there exists an abelian tower for $G$ in which all subgroups in the tower are normal in $G$.
For a counterexample to the statement you seem to be trying to prove, start with the Klein $4$-group, $C_2\times C_2$, written as $\{1, x, y, xy\}$ (that is, $x\leftrightarrow (g,e)$, $y\leftrightarrow(e,g)$, and $xy\leftrightarrow (g,g)$, where $C_2$ is generated by $g$); let $G=(C_2\times C_2)\rtimes C_3$, with $C_3$ generated by an element $z$ and $z$ acting on $C_2\times C_2$ by sending $x\mapsto xy\mapsto y\mapsto x$.
Then $G$ has an abelian series given by $$\{1\} \subseteq \{1,x\}\subseteq \{1,x,y,xy\}\subseteq G.$$ If we let $G_0=G$, $G_1=\{1.x,y,xy\}$, $G_2=\{1,x\}$, and $G_3=\{1\}$, note that $G_2$ is abelian; that $G_2$ is normal in $G_1$ because $G_1$ is abelian, and the quotient is cyclic of order $2$. $G_1$ is normal in $G_0$ by construction, and the quotient is cyclic of order $3$. However, $G_2$ is not normal in $G$, since conjugating $x$ by $z$ gives $y$, which is not in $G_2$.
So you cannot hope to prove the result as you are attempting to.
The standard proof uses the commutator subgroup, and the derived series. Recall that if $G$ is a group, its commutator subgroup $G'$ (also denoted $[G,G]$) is defined to be $$G' =[G,G] = \langle [x,y]\mid x,y\in G\rangle,$$ where $[x,y]=x^{-1}y^{-1}xy$. It is a good exercise to check that:
Define the derived series of $G$ by: $$\begin{align*} G' =G^{(1)} & = [G,G];\\ G'' = (G')' = G^{(2)} &= [G',G'];\\ G^{(n+1)} = (G^{(n)})' &= [G^{(n)},G^{(n)}]. \end{align*}$$ Note that $G^{(k)}\triangleleft G$ for all $k$, thanks to point 1 above; because if $H$ is normal in $G$ and $K$ is characteristic in $H$, then $K$ is normal in $G$.
Now one proves the desired result, that if $G$ is solvable then it has an abelian tower in which every group that occurs in the tower is normal in $G$ as follows: Suppose that $$1 = G_m \leq G_{m-1} \leq \cdots \leq G_1 \leq G$$ is a normal tower in which $G_i/G_{i+1}$ is abelian. Then $G'\leq G_1$; therefore, $G'' = (G')'\leq (G_1)'$. But since $G_1/G_2$ is abelian, $(G_1)'\leq G_2$, so $G''\leq G_2$.
Assume we have proven that $G^{(k)}\leq G_k$. Then $G^{(k+1)}\leq G_k'$. And since $G_k/G_{k+1}$ is abelian, then $G_k'\leq G_{k+1}$, so $G^{(k+1)}\leq G_{k+1}$.
Inductively, we conclude that $G^{(m)}\leq G_m=\{1\}$, so $G^{(m)}$ is trivial. Therefore, we have an abelian tower with $$1 = G^{(m)}\leq G^{(m-1)}\leq \cdots \leq G' \leq G,$$ and all terms are normal in $G$, as desired. $\Box$