I must prove that in a non-separable graph G, (non-separable: the removal of any vertex of G will result in a disconnected graph) let k= the maximum length of a cycle in G, and let C and C' both be k-cycles. Then C and C' have at least 2 vertices in common.
Now, I understand why this must be so. If there are no vertices, or one vertex in common then I either have that G is separable (a contradiction) or that there exists a cycle with length greater than k (a contradiction). However, If C and C' have two vertices in common, let's call them U0 and V0, and let all the vertices of C be called Ui for all 1<=i<=k. and let the names of the vertices in C' be Vj for all 1<=j<=k. then I construct the cycle U0 to U1 to U2 to U3 to U4 to ... to V0 to V1 to V2 to V3 ... to U0. then I have a cycle of length greater than k... thus we have a contradiction in every case where C and C' have common vertices except in the case where C=C'. The conclusion that I'm reaching is that a cycle of maximum length in a non-separable graph G must be unique. I don't think this is the conclusion I'm supposed to reach. Could anyone offer a suggestion of where I am going wrong? Or is this in fact right: we can only have 1 cycle of maximum length in a non-separable graph G.?