Sequence of elements representing a Hamiltonian cycle and the generating element of a subgroup

Question

Let $S$ be a subset of a finite group $G$. The Cayley graph $Cay(G,S)$ can be defined as the graph whose vertices are the elements of $G$, with an edge joining $g$ and $gs$, for every $g \in G$ and $s \in S$.

For $s_1,s_2,\cdots,s_n \in S \cup S^{-1}$, $(s_1,s_2, \cdots, s_n)$ is used to denote a walk in $Cay(G,S)$ that visits (in order) the vertices, $e, s_1, s_1 s_2, \cdots, s_1 s_2 s_3 \cdots s_n$.

Lemma: Suppose,

1. $S$ is a generating set of $G$,

2. $H$ is a cyclic subgroup of $G$, with index $|G:H|=n$,

3. $s_1,s_2,\cdots,s_n$ is a sequence of $n$ elements of $S \cup S^{-1}$ such that,

i) the elements $e, s_1, s_1 s_2, \cdots, s_1 s_2 s_3 \cdots s_n$ are all in different right cosets of $H$,

ii) the product $s_1 s_2 \cdots s_n$ is a generator of $H$.

Then, $(s_1, s_2, \cdots , s_n)^{|H|}$ is a Hamiltonian cycle in $Cay(G,S)$.

In this lemma, what is the importance of having $s_1 s_2 \cdots s_n$ as a generator of $H$?

I have seen other lemmas and theorems too where similar condition that such a product should be a generating element of some subgroup is mentioned. But I'm unable to see why?

Also, suppose a case where $S=\{u,t\}$. Then a Hamiltonian cycle in a Cayley graph with respect to $S$ will be expressed in terms of $u,t$. Then, if the Hamiltonian cycle is $(u,t,t^{-1},t, \cdots, t)^{|H|}$, I thought probably the above definition of notations mean to take $s_1=s, s_2=t, s_3=t^{-1}, s_4=t, \cdots, s_n=t$ since they have not mentioned that $s_1, s_2, s_3, \cdots, s_n$ are distinct elements. Am I right?.

Can some one help me with the above questions.

Thanks a lot in advance.

Answer 1

In the walk $(s_1, s_2, \dots , s_n)^{|H|}$, we visit $H$ every $n$ elements, so after $n|H|$ steps, the elements of $H$ we see are exactly $(s_1 s_2 \cdots s_n)^k$ for $0 \le k \le |H|-1$. If $s_1 s_2 \cdots s_n$ does not generate $H$, then we won't get a Hamiltonian cycle, because we won't visit all the different elements of $H$.

And yes, it's fine for some of the elements $s_1, s_2, \dots, s_n$ to coincide in the definition of $(s_1, s_2, \dots, s_n)$.