For every 3-regular graph $G$ there exists a 3-regular planar graph $G'$ such that $C(G) = C(G')$

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$C(G)$ denotes the cycle set of $G$ (the set of lengths of all cycles of $G$). This question came up recently and I have been meaning to take a stab at it, but I wanted to know if there exists some well known counterexample. Even dropping 3-regularity, the statement is not at all trivial. For example, thinking that one could simply construct such a graph $G'$ by

$$ G' = \bigoplus_{n \in C(G)} \mathcal{C}_n $$

Where $\mathcal{C}_n$ denotes the cycle graph on $n$ vertices is simply wrong, since this construction introduces more cycles in $G'$ that are not in $G$. What is perhaps more interesting, is that this statement seems to imply the Erdős–Gyárfás conjecture, if paired with the main result of this paper. Could this ever possibly be true?

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The Heawood graph is a counterexample.

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It is a $3$-regular graph with girth $6$; the set of cycle lengths in the graph is $\{6, 8, 10, 12, 14\}$.

A $3$-regular planar graph cannot have girth $6$. This is a bound via Euler's formula: in general, if the girth of an $n$-vertex planar graph is $k$, it has at most $\frac{k}{k-2}(n-2)$ edges, which is at most $\frac64(n-2) = \frac32n - 3$ edges in our case. But a $3$-regular $n$-vertex graph must have exactly $\frac32n$ edges.