Show that for any two edges of a 2-connected graph, a cycle exists containing both of them.

Here's what I have tried so far:

Lemma. A graph $G$ is 2-connected if and only if every two vertices are contained in a cycle.

Attempt: Let $e_1=\{v_1,v_2\}$, $e_2=\{w_1,w_2\}$ be two edges of $G$. By lemma, there is a cycle $C_1$ containing $v_1,w_1$ and a cycle $C_2$ containing $v_2,w_2$. Denote the two paths from $v_1$ to $w_1$ in $C_1$ as $p_1$ and $p_2$ and similarly set $r_1$, $r_2$ for the two paths in $C_2$. Then divide the problem into different cases:

Case 1: Neither $p_1$ nor $p_2$ intersects $r_1$ or $r_2$.

Case 2: Only one of $p_1$ and $p_2$ intersects $r_1$ or $r_2$.

Case 3: Both intersect $r_1$ or $r_2$.

For the first two cases it is easy to show that such cycle exists by just picking one path from each $C_1$ and $C_2$. But the third case is a lot complicated so I believe there should be a much nicer way to approach this problem. Any idea?



  • If $e$ belongs to any cycle, then subdividing it (creating a vertex in the middle) doesn't break 2-connectivity.

I hope this helps $\ddot\smile$

  • $\begingroup$ sorry I am still unsure how to proceed with this. Why would I subdivide one edge? Thank you very much. $\endgroup$
    – user112358
    Oct 25 '16 at 10:03
  • $\begingroup$ @Lewis You asked about a nicer way to prove your claim. $\endgroup$
    – dtldarek
    Oct 25 '16 at 11:19

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