Proof that a given commutator subgroup is a subgroup of another group I am trying to understand the proof of the following lemma.

Lemma: The commutator subgroup of $F_{i+1}$ is a subgroup of the group $\langle x_1, \ldots, x_{i-1}\rangle = g^{-1}F_{i-1}g$, where
   $(1 \leq i \leq t-2)$.
Proof: Provided that $1 \leq i \leq t-2$, the groups $F_i = \langle x_0, \ldots, x_{i-1} \rangle$ and $g^{-1}F_ig = \langle x_1,
 \ldots, x_i \rangle$ are different, and they are both of index two in
   $F_{i+1}$, and consequently normal in $F_{i+1}$. Thus their
   intersection $\langle x_1,\ldots,x_{i-1} \rangle = g^{-1}F_{i-1}g$ is
   normal in $F_{i+1}$, and the quotient group $F_{i+1}/(g^{-1}F_{i-1}g)$
   is Abelian, since it has order 4. Hence the commutator subgroup of
   $F_{i+1}$ is contained in $g^{-1}F_{i-1}g$.

The proof is from "Algebraic Graph Theory" by Biggs (1974), and is shown on page 123 (if anybody happens to have the book). Although the subject is algebraic graph theory, I am quite sure that the proof relies on regular group theory.
Edit: Here are some more details, which in retrospect were important
First, the group $F_i$ is defined as: $F_i = \langle x_0, x_1, \ldots, x_{i-1} \rangle$. There is the following relation between $g$ and $x_i$
$$x_i = g^{-i}x_0g^i$$
The elements $x_i$ are involutions, that is $x_i = x_i^{-1}$. The order $|F_i| = 2^i$. This (as mentioned by Jack, this gives immediately that the index of $F_i$ in $F_{i+1}$ is 2, since $|F_{i+1}/F_i| = |F_{i+1}|/|F_i|$. The group $F_0$ is the identity.
Current status: At this point, I am almost through the proof, and currently the only thing which I do not understand is the sentence:

Provided that $1 \leq i \leq t-2$, the groups $F_i = \langle x_0, \ldots, x_{i-1} \rangle$ and $g^{-1}F_ig = \langle x_1,
 \ldots, x_i \rangle$ are different

I do not see why this is true, and I do not see why it is important. If somebody can answer that sub-question, I will mark the answer as correct.
 A: While this post has been here, I have figured out the last details of the proof. Below is my new expanded proof:

Provided that $1 \leq i \leq t-2$, the groups $F_i = \langle x_0, \ldots, x_{i-1} \rangle$ and $g^{-1}F_ig = \langle x_1, \ldots, x_i \rangle$ are different. This is true, since if they were equal, the group $F_i$ would include $x_i$, which would make it equal to $F_{i+1}$, but we know that the order of $F_i$ and $F_{i+1}$ is not the same.
Since $|F_i| = 2^i$, we have that
  $
|F_{i+1} / F_i| = \frac{|F_{i+1}|}{|F_i|} = \frac{2^{i+1}}{2^i} = 2
$
  and the same holds for $g^{-1}F_ig$. Thus both $F_i$ and $g^{-1}F_ig$ are index 2 in $F_{i+1}$. Since a subgroup of index 2 is normal, we then have that $F_i$ and $g^{-1}F_ig$ are normal subgroups in $F_{i+1}$. Since the intersection of two normal subgroups is also a normal subgroup, this gives us, that their intersection $\langle x_1,\ldots,x_{i-1} \rangle = g^{-1}F_{i-1}g$ is normal in $F_{i+1}$. Again using the order of $F_i$, we see that the quotient group $F_{i+1}/(g^{-1}F_{i-1}g)$ has order 4, and is therefore Abelian. At last, we use that for a group $G$, the quotient group $G/N$ is Abelian, iff the commutator subgroup of $G$ is a subset of N. This gives that $F_{i+1}/(g^{-1}F_{i-1}g)$ is Abelian -- and shows that the commutator subgroup of $F_{i+1}$ is contained in $g^{-1}F_{i-1}g$.

