I'm currently reading a paper on finding Hamiltonian paths in Cayley graphs and the author makes a claim that I can't seem to understand.
Let $G$ be a finite, nilpotent group, $N$ a normal subgroup that contains $[G, G]$ with $|N|$ square-free, and $S = \{\sigma_1, \dots, \sigma_\ell\}$ a subset of $G$ such that $\overline S$ is a minimal generating set for $\overline G$, where $\overline H$ denotes the image of $H$ in $G / N$. We have the following notation:
$$ S_k = \{\sigma_1, \dots, \sigma_k\} $$ $$ G_k = \langle S_k \rangle N $$ $$ m_k = |\overline {G_k} : \overline {G_{k-1}}|.$$
The author makes the following claim: There exists an integer $r$, $0< r \leq |\overline \sigma_1|$, such that $$ \overline \sigma_2^{m_2} = \overline \sigma_1^r. $$
I suppose my main issue is that I don't have an intuitive understanding of what $m_2$ is supposed to represent. Thank you very much!
$m_2$ is the order of the quotient group $H := \langle \overline{\sigma_1}, \overline{\sigma_2} \rangle / \langle \overline{\sigma_1} \rangle = \overline{G_2}/\overline{G_1}$ (notice that $\overline{G_1} \unlhd G/N$, since $G/N$ is abelian, as $[G, G] \le N$). Thus the order of $\overline{\sigma_2}$ in $H$ divides $m_2$, so $\overline{\sigma_2}^{m_2} \in \langle \overline{\sigma_1} \rangle$. This says precisely that $\overline{\sigma_2}^{m_2} = \overline{\sigma_1}^r$ for some $0 < r \le |\overline{\sigma_1}|$.
For this argument, we don't need $G$ to be nilpotent, or $|N|$ squarefree (also, $[G,G] \le N$ will imply $N$ is normal).