# How big can the girth of the a graph $G$ consisting of two $C_k$ + some edges to make $G$ cubic?

Let $$G$$ be a simple graph with vertex set $$\{u_1, \dots, u_n\} \cup \{v_1, \dots v_n\}$$, with $$G_{\uparrow U}$$ (subgraph induced by $$U$$) and $$G_{\uparrow V}$$ being cycles of size $$n$$. Now we want to add edges between each vertex of $$U$$ to a unique vertex of $$V$$ such that the length of the minimum cycle (girth) is as big as possible.

Question: Is it possible to have a girth of $$\Omega(\log \log n)$$ for infinitely values of $$n$$? If not, was girth lower bound can asymptotically achieve on such graphs?

An easy edge set would be $$\{u_i, v_{i+k} \,$$mod $$n\}$$ but it gives cycles of size 4 almost everywhere.

I have a feeling the answer could be related to $$\mathbb{Z}/p \mathbb{Z}$$ groups as other simple construction could be the edge set "{$$u_x, v_y$$ | $$y \equiv px$$ mod $$n\}$$" with probably $$p$$ and $$n$$ being a prime numbers (I talk with 8+ years memory from my arithmetic classes).

However I can see how studying these groups would help to find cycles whose intersection with $$G_{\uparrow U}$$ is a single path, but not for other type of cycles.

this paper shows that for all girth $$g$$ there such a graph. Moreover it gives a recursive construction on $$g$$: if one can build such a graph with $$n$$ nodes of girth $$g$$, then there is a graph on $$O(n 2^n)$$ nodes with girth $$g+1$$.