Suppose we have two primitive cycles $C_1$ and $C_2$ which belong to two different equivalence classes.
Is the cycle $C_1 C_2$ also primitive?
My intuition is that this cycle is also primitive, but I fail to compose a proof—I could not find a direct contradiction by making the proposition false.
This question is raised in the context of the zeta function of a finite graph, and primitive cycles have the usual meaning.
In case that terminology should be fixed, related definitions are listed below.
- A walk is just represented by a sequence of arcs (directed edges).
- A walk is closed if it starts and ends with the same vertex.
- A walk is non-backtracking if it does not have a subsequence like $\{i \to j, j \to i\}$.
- A cycle is a walk which is closed and non-backtracking.
- A cycle is primitive if it is not obtained by repeating another cycle $m>1$ times.
- Two cycles are said to be equivalent if one can be obtained by the cyclic permutation (i.e., reselecting the starting vertex) of another.
- By this equivalence relation, cycles can be classified into different equivalence classes.
In particular, primitive cycles can be classified into classes and these classes of primitive cycles are used to define the zeta function of a finite graph.
Best Answer
How about: \begin{align*} C_1\colon 0\to a\to b\to 0 \to c\to d\to 0\to a\to b\to 0 \\ C_2\colon 0 \to c\to d\to 0\to a\to b\to 0 \to c\to d\to 0 \end{align*}