Let $X$ be a trivalent graph. We denote by ${\rm Aut}_e(X)$ the subgroup of ${\rm Aut}(X)$ such as fix the edge $e$,
i.e., $\forall \phi \in{\rm Aut}_e(X)$ if $e = (v_1, v_2)$ then $ \phi(v_1) = v_2$ and $ \phi(v_2) = v_1$
or $( \phi(v_1) = v_1$ and $ \phi(v_2) = v_2$.
${\rm Aut}_e(X_r)$ where $X_r$ is the subgraph consisting of all vertices and all edges of the graph$X$ which appear in paths of length $\leq r$ through $e = (a, b)$.
There are natural homomorphisms:
$$\pi_r :{\rm Aut}_e(X_{r+1}) \rightarrow{\rm Aut}_e(X_r)$$ in which $\pi_r(\sigma)$ is the restriction of $\sigma$ to $X_r$.
If $K_r$ is the kernel of $\pi_r$, and $\pi_r({\rm Aut}_e(X_{r+1}))={\rm Im} \; \pi_r$.
Given that:
(1) $|{\rm Aut}_e(X_{r+1})|=|{\rm Im}\; \pi_r| \cdot |K_r|$,
(2) $K_r$ is the elementary abelian 2−group.
I have to prove using induction :
${\rm Aut}_e(X_{r+1})$ is a 2−group.
I understand why $K_r$ is the elementary abelian 2−group but not clear how the following equation is found:
$|{\rm Aut}_e(X_{r+1})|=|{\rm Im}\; \pi_r| \cdot |K_r|$
and how to run induction to prove ${\rm Aut}_e(X_{r+1})$ 2−group. Please, help about above 2 problems.
Thanks.
EDIT:
$|{\rm Aut}_e(X_{r+1})|=|{\rm Im}\; \pi_r| \cdot |K_r|$ could be found by using Lagrange theorem, where it is evident that $K_r$ is a subgroup of ${\rm Aut}_e(X_{r+1})$, then ${\rm Im}\; \pi_r$ has to be the set of cost representative. In that case ${\rm Aut}_e(X_{r})= {\rm Aut}_e(X_{r+1})/K_r$? Is it correct?