Graph theory: cycles

Prove that if two distinct cycles of a graph G each contain an edge e, then G has a cycle that does not contain e.

My approach is since they both have edge e then if we remove edge e from both then connect the two cycles together at vertex a and b which edge e connected to then we would get a cycle. Am I missing something here?

This isn't a homework question, just a question from my textbook.

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Another approach would be to take the connected component that contains both cycles and observe that in this subgraph $|E| \geq |V|+1$. To be more explicit:

• $|E| \leq |V|-2$ the graph would not be connected,
• $|E| = |V|-1$ the graph would be a tree (no cycles),
• $|E| = |V|$ the graph would have exactly one cycle,
• $|E| \geq |V| + 1$ is the only possibility.

Then if we remove $e$, we have $|E - \{e\}| \geq |V|$ which still implies the existence of some cycle, which, of course, cannot contain $e$ (it is not necessary, but if you wonder, this connected component is still connected because $e$ was an edge of a cycle).

I hope this helps ;-)

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