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.