Reconciling Different Definitions of Solvable Group

Looks like my class note defines solvable group differently from others:

A finite group $G$ is called solvable if $H' \neq H$ for each subgroup $H$ of $G$ different from $\{1\}$,

where $H'$ is called the commutator subgroup of $G$ (correct me if I am wrong): $$\begin{align} H' :&= [H, H] \\ &= \langle [a, b] \mid a, b \in H \rangle \\ &= \langle a^{-1}b^{-1}ab \rangle. \end{align}$$

And then on the same text there is problem like this:

Show that a finite group $G$ is solvable if and only if these two are met:
$(\mathscr C_1): \{1\} = H_0 \lhd H_1 \lhd \ldots \lhd H_{i-1} \lhd H_i \lhd \ldots \lhd H_n = G$, where $H_i < G,$ and $i =\{1, 2, \ldots n\}$;
$(\mathscr C_2):$ Factor group $H_i/H_{i-1}$ is abelian.
(Note that these $\mathscr C$'s are actually Wikipedia's definition of solvable group here.)

I think I am done proving that if $\mathscr C$'s are true $\Rightarrow G$ is solvable, but I am struggling with proving that if $G$ is solvable $\Rightarrow \mathscr C$'s. Any hints or help would be appreciated, thanks for your time.

POST SCRIPT: ~~~~~~~~~~~~~~~~~~~~~~~~~~

I have been working on this problem since first posted and, after getting some help from “Modded Bear” here, I am able to put together the first part ofsolution like these:

Lemma: Suppose that $G$ is solvable, then (i) each subgroup of $G$ is solvable, and (ii) each factor group of $G$ is solvable.

(A) Proving that if $\mathscr C$'s are true then $G$ is solvable:
(1) Since $\mathscr C_2: H_i/H_{i-1}$ is abelian, therefore $\forall a, b \in H_i$, we have

$$\begin{align} (H_{i-1})a(H_{i-1})b &= (H_{i-1})b(H_{i-1})a \tag{1}\\ (H_{i-1})ab &= (H_{i-1})ba \tag{2}\\ ab(ba)^{-1}(H_{i-1}) &= (H_{i-1}) \tag{3}\\ \underbrace{aba^{-1}b^{-1}}_{\in \ H'_{i}} &\in (H_{i-1}) \tag{4}\\ \therefore \forall x \in H'_{i} &\rightarrow x \in H_{i-1} \tag{5}\\ H'_{i} &< H_{i-1} \tag{6}\\ \because H_{i-1} < H_{i} &\rightarrow H'_{i-1} < H'_{i}\tag{7}\\ \therefore H'_{i-1} &\neq H_{i-1} \tag{8} \end{align}$$ (2) With similar analysis, we can easily derive $H'_{i} \neq H_{i}$ for each subgroup of $G$, therefore per definition $G$ is solvable as desired. $\blacksquare$

(B) Proving that if $G$ is solvable then $\mathscr C$'s are true:

(1) ...
(2) ...

For part (b), the goal is to show that if $G$ is a finite group with the property that $H' < H$ for every subgroup $H$, then there is a subnormal series $$1 = H_0 \lhd H_1 \lhd \cdots \lhd H_n = G$$ where the factor groups $H_{i+1}/H_i$ are all abelian. (This is one definition of a solvable group.)

To get started, try setting $H_{n-1} = G'$. Then $G' < G$ (because of the given property) and $G' \lhd G$ (in fact, $G'$ is a characteristic subgroup of $G$), so $H_{n-1} \lhd H_n$ and therefore $H_n / H_{n-1}$ is a group.

To show that $H_n / H_{n-1}$ is abelian, we need to show that any two elements $aH_{n-1}$ and $bH_{n-1}$ commute, where $a$ and $b$ are elements of $H_{n}$. So we require $$aH_{n-1} bH_{n-1} = bH_{n-1} aH_{n-1}$$

This will be true if and only if $$abH_{n-1} = baH_{n-1}$$ if and only if $$a^{-1}b^{-1}abH_{n-1} = H_{n-1}$$ if and only if $$a^{-1}b^{-1}ab \in H_{n-1}$$ and this is of course true because $H_{n-1}$ contains the commutators of all of the elements in $H_n$. We conclude that $H_n / H_{n-1}$ is abelian.

So that takes care of the first step in the subnormal series. If $H_{n-1}$ happens to be $1$, then we're done. Otherwise, you need to keep repeating this process: let $H_{n-2} = H_{n-1}'$. Once again, $H_{n-1}' < H_{n-1}$ (given property), and once again you want to show that $H_{n-1}/H_{n-2}$ is abelian. The exact same argument as above will work.

Eventually this process must terminate, because each subgroup is strictly smaller than the previous one, and $G$ is finite, so eventually one of the subgroups will have to be $1$.