Context
Consider the following Theorem from a lecture about algorithmic group theory, I'm taking at the moment:
Let $N \not =\{1\}$ be a finite generated normal subgroup of a free group $F_A$ with $A$ being the alphabet, and let $H\leq F_A$ be a finitely generated subgroup of $F_A$ with $N\leq H$. Then the Stallings automaton $\Gamma(H)$ for the subgroup $H$ is complete.
I'm trying to understand the following proof:
We proof, that for any $a \in A \cup \overline{A}$ we can find a reduced word $v_a = av \in N$:
Let $w$ be a generator of $N$, then write $w$ as reduced word and $w = uv\overline{u}$ with $v$ being cyclic reduced. Now add $v$ to the set of generators and consider and $a \in A \cup \overline{A}$.
- If $a$ apperars in $v$, then $v_a=v'av''$. The word $av''v' \in N$ is reduced
- If $\overline{a}$ appears in $v$, then add $\overline{v}$ to the generators and considere $\overline{v}_a$
- If neither $a$ nor $\overline{a}$ appear in $v$, then $v_a=av\overline{a} \in N$ is reduced
For every word $uw \in H$ also $uv_aw \in H$, because $N$ is a normal Subgroup in $H$: $$uv_aw = uw\overline{w}v_aw \in H\cdot N \subset H$$ Now stick all words $v_a$ to every node as petal. This does not produce nodes of degree 1. Now every petal can be stallingsreduced completly, which proves that $\Gamma(H)$ has been complete already.
My question
I get the basic principle of the proof, tyring to complete $\Gamma(H)$ just for showing that this is redundant. But what seems strange to me, is the way we seperate cases for $v_a$. The last case for generating $v_a$ seems to be the most straightforward to me and I understand that this case does not work if $v$ starts with $\overline{a}$ or ends with $a$ because it isn't reduced and we therefore need the other two cases. But why are we interested in $a, \overline{a}$ appering anywhere in $v$, I don't see a problem with the last case when $a$ or $\overline{a}$ is contained in any other place of $v$.
So, is there a specific reason, why the first two cases are more general than needed?