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) ...
 A: 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 $1$.
A: The idea for the other direction is to pick some natural $H_i$s guaranteeing the factor groups are abelian. $G'$ seems tempting for $H_{n-1}$. Do you see what to pick next, and how solvability guarantees you'll make it down to $e$ eventually?
