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Show that if a simple graph with at least two vertices is connected and has no cut vertices, then any two vertices lie on a cycle and any two edges lie on a cycle.

I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?

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  • $\begingroup$ What have you tried so far? $\endgroup$ – jwc845 Nov 19 '18 at 15:37
  • $\begingroup$ @jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle? $\endgroup$ – Thomas Nov 19 '18 at 15:58
  • $\begingroup$ This might help: math.stackexchange.com/questions/3005438/… $\endgroup$ – Just_a_newbie Nov 24 '18 at 10:44

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