Basic Questions in Graph Theory Let G be our not directed graph and each vertex have degree(2) prove that there is  a cycle in the graph.
Can somebody explain me how to solve problem of that type?
Any help will be apprecited!
 A: We will show that an undirected finite graph $G$, in which each vertex has degree at least $2$, contains a cycle. Choose any vertex as your starting point, and start walking along the edges of the graph, without using any edge twice. Note that, whenever you arrive at a new vertex (not visited previously), you will always be able to leave it, because of the degree condition. Since the graph is finite, you can't go on visiting new vertices forever; eventually some vertex $v$ will be repeated. The portion of your walk, between your first and second visit to $v$, is a cycle in $G$.
A: Let $(v_1, v_2, \dots, v_n)$ in $G$ each have degree 2 with the exception of the first and last vertex $v_1$ and $v_n$ respectively, which has degree 1.
Then there is at least an (Euler) path from $v_1$ to $v_n$ in $G$.
But then adding an edge between each vertex crossed from $v_1$ to $v_n$ we get a backwards path.
Therefore there is a Euler circuit from $v_i$ to $v_j$ (a circuit - or cycle - that uses each edge only once). QED
A: You can prove a slightly stronger claim. If $G$ is a (multi)graph with minimal degree $2$ then $G$ contains a cycle.
The proof is easily done by induction on the number of vertices of $G$. Assuming you know how to do the base case, suppose $G$ is a graph with $n+1$ vertices and minimal degree $2.$ By contracting an edge you obtain a (multi)graph $G'$ of order $n$ and minimal degree $2$ hence by the induction hypothesis it must contain a cycle. It is now very natural to extend this cycle of $G'$ to a cycle in $G.$
