Is the cycle graph of a group unique? I was perusing the cycle graphs for small groups on Wikipedia and something bothers me: is the cycle graph of a finite group actually unique (up to isomorphism)?
For example, if there are any cyclic subgroups of order $5$, the cycle graph is drawn by picking one primitive generating element $a$, and drawing a $5$-cycle in the graph between $e, a, a^2, a^3, a^4$. But a priori, this means the graph will depend on the choice of $a$. Can this result in multiple cycle graphs for the same group? It doesn't seem obvious that these different graphs would be isomorphic in general.

Definition: A cycle graph of a finite group $G$ is a simple undirected graph defined as follows: first, the vertex set of the graph is taken to be the set of elements $g \in G$. Then, for each maximal cyclic subgroup of $G$ (cyclic subgroup not fully contained in a larger cyclic subgroup), pick a generator $a$ of the subgroup, and draw undirected edges $e \to a \to a^2 \to a^3 \to \cdots \to a^{k-1} \to a^k = e$ (ignoring any duplicate edges), where $k$ is the order of the subgroup.
My question is whether the cycle graph of $G$ is unique up to isomorphism, regardless of the choices of generator for each maximal cyclic subgroup. Notice that for the purposes of this question, the graph is completely unlabeled -- the original vertex labels (elements of the group) are ignored, and edges are not labeled with the cyclic subgroup they correspond to.

Strangely, I can't find a previous thread on this:

*

*Do cycle graphs determine groups up to isomorphism? asks the converse question of whether the cycle graph uniquely determines the group; Chris Cutler asks my question in the comments but is unanswered.


*How in general does one construct a cycle graph for a group? asks for how to construct the cycle graph, but the top answer suffers from the same problem that the choice of primitive element for a cycle is not unique.


*I also searched on Google Scholar. I found an interesting paper, The Cyclic Graph of a Finite Group (Ma, Wei, Zhong), but it defines the cycle graph differently, where $x, y$ share an edge if $\langle x, y \rangle$ is cyclic. In this definition the graph is clearly unique. This also seems to me a much more sensible definition, but I don't have an example where Wikipedia's definition actually leads to ambiguity in the resulting graph, up to isomorphism.
 A: In general, the isomorphism type of the cycle graph does depend on the choices of generators. Here is an example (inspired by Hagen von Eitzen's comment).
Let $G=C_{10}\times C_2$, with the first factor generated by $a$ and the second generated by $b$. This group has three maximal subgroups, all of order $10$. So the edges of the cycle graph can be decomposed into three $10$-cycles. The five elements in the characteristic subgroup  $H$ of order $5$ are contained in every subgroup of order $10$, so they have valency $6$, while the elements of order $2$ and $10$ are contained in a unique subgroup of order $10$, so they have valency $2$.
Now, for some choices of generators, for example $\{a,ab,a^6b\}$, elements of $H$ have exactly two elements at distance $2$, which happen to be in $H$, and there are three $2$-paths between them, as in
https://en.wikipedia.org/wiki/File:GroupDiagramMiniC2C10.png
For some other choices of generators, for example $\{a,ab,a^2b\}$, this is not the case. (In this case, there are four vertices at distance $2$, some with a unique path, some with two paths.)
A: For completeness, here are drawings of the graphs that are discussed in verret's answer.
Both of these are valid "cycle graphs" for the abelian group $C_{10} \times C_2$, otherwise known as $C_5 \times C_2 \times C_2$ or, as an additive group,
$$
(\mathbb{Z}\;/\;5\mathbb{Z})
\times (\mathbb{Z}\;/\;2\mathbb{Z})
\times (\mathbb{Z}\;/\;2\mathbb{Z})
$$
The red elements are the central subgroup $\{(a, 0, 0) \;\mid\; a \in \mathbb{Z} \;/\; 5\mathbb{Z}\}$, which is equal to the intersection of all three maximal subgroups. The identity element is unlabeled, but assume it is at the top. First, using generators $(1, 1, 0), (1, 1, 1), (1, 0, 1)$, we have that all three generators double to get $(2, 0, 0)$, so we have the following graph:

Alternatively, using generators $(1, 1, 0), (1, 1, 1), (2, 0, 1)$, the first two generators double to get $(2, 0, 0)$ but the third doubles to get $(4, 0, 0)$ instead. We end up with the following graph:

Finally, the graphs are non-isomorphic because the first graph has diameter 5, while the second has diameter 4.
