I thought of proving it first for the cases where G is connected (so I would prove that G is a cycle), and later consider the cases where G is not connected and apply this result to each of the connected components of G, obtaining that G is a disjoint union of cycles.

If $\delta=2$, I could prove this by picking a vertex $v_1\in V(G)$ and, since $\delta=2$, pick $v_2\in V(G)$ such that it is one of the neighbors of $v_1$. Then we pick $v_3$ such that is a neighbor of $v_2$ and $v_3\neq v_1$. We iterate this process until we pick a vertex $v_n$ neighbor of $v_{n-1}$ such that $v_n\neq v_i$, $i=1,...,n-1$, so we have a path $L=v_1v_2.....v_{n-1}v_n$ of lenght $n-1$. Because $\delta=2$, $v_n$ must have another neighbor and the only option left is $v_1$. Thus, we get that G is a cycle.

However, since $\delta\ge2$, along this process I could be choosing a vertex $v_i$ that not only is neighbor of $v_{i-1}$ and $v_{i+1}$, but also of a vertex $v_j$, where $j=1,...,i-2$.


1 Answer 1


The total of the valencies of the vertices in a graph with $e$ edges is $2e$, and if the graph has $n$ vertices the mean valency is $2e/n$. Here the $e=n$ and so the mean valency is $2$, but also all vertices have valency $\ge2$ so every vertex has valency exactly two.

So at each vertex one can start a walk, leaving on one edge, and at each vertex leaving from the different edge to the one one arrived on, until one eventually reaches the start vertex, completing a cycle.


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