Prove that every vertex of a 2-regular graph G lies on an exactly one circle My only idea so far is that if a vertex $v$ would lie on more than one circle, i. e. on 2 circles, then those 2 circles must separate in some vertex in order to be different and that cannot be because each vertex has  a degree of 2.
I'm not even sure whether my assumption could somehow help me in proving the statement. I would be thankful for some hints.
 A: It suffices to prove that every 2-regular graph the disjoint union of cycles. Let $G$ be a 2-regular graph. Pick any vertex $v_1$ of $G$.  The vertex $v_1$ has at least two neighbors.  Say $v_2$ is a neighbor.  Now $v_2$ has two neighbors, and so has another neighbor besides $v_1$. Call this neighbor $v_3$.  By repeating this process, we get a sequence of vertices $v_1, v_2, \ldots$.  Since the graph is finite, at some point we repeat a vertex, $v_k$ say.  So $v_k, v_{k+1}, \ldots,  v_k$ is a cycle $C$ in the graph $G$.  Observe that $C$ is a 2-regular subgraph of $G$. Because $G$ is 2-regular, the edges of $G$ incident to each vertex of $C$ are also the edges of $C$. We remove from $G$ the edges and vertices of $C$. The induced subgraph $G-C$ is again a 2-regular graph.  We now repeat the above procedure to get another cycle, and so on.
A: Here is another answer:
Since $G$ is $2$-regular, sum of vertices degrees of $G$ is $2n$ hence $m = n$. Now consider $G$ and suppose it has $r$ different connected components. Call them $G_1$ through $G_r$ and suppose $G_i$ has $k_i$ vertices. We have $k_1+k_2+...+k_r = n$. Suppose $T_i$ is a spanning tree of $G_i$ and hence has $k_i-1$ edges so $T_1$ through $T_r$ omit $r$ edges of $G$. As $G$ is $2$-regular, each $T_i$ is a path so is $P_{k_i}$. (Root each tree from a leaf and assertion becomes evident) Any edge between non-endpoint vertices of $P_{k_i}$ results in a vertex of degree $3$ so the only possible edge that can be added is the one between the endpoints of the path and as each component $G_i$ must have $k_i$ edges each such edge should be present hence we conclude each $G_i$ is isomorphic to $C_{k_i}$ and we conclude a stronger result other than each vertex being present on exactly one cycle. $\blacksquare$
