# Prove that if $G$ is not 2-connected then there exists a pair of edges such that no cycle contains both edges

Full question: Prove that if $$G$$ is loopless, has no isolated vertices, has at least $$2$$ edges and is not $$2$$-connected then there exists a pair of edges $$\{e_1, e_2\}$$ in $$G$$ such that no cycle contains both $$e_1$$ and $$e_2$$.

So far I have that $$G$$ is not $$2$$-connected $$\therefore$$ must contain proper separations of order $$0$$ or order $$1$$. If $$G$$ contains only proper separations of order $$1$$ then $$G$$ is connected. If $$G$$ contains a proper separation of order $$0$$ then $$G$$ is disconnected. Then I have carried on from there. However the proof ends up quite verbose so I was wondering if there was a more simple way to do it.

If $$G$$ is not connected then let $$e_1$$ be an edge in one component $$C_1$$ of $$G$$ and let $$e_2$$ be an edge in another component $$C_2$$ of $$G$$. Then clearly there is no cycle containing both $$e_1$$ and $$e_2$$.
Let us assume now that $$G$$ is connected but not 2-connected. Then let $$w$$ be such that $$G \setminus \{w\}$$ is not a connected graph. Then let $$C_1$$ be one component of $$G \setminus \{w\}$$ and let $$C_2$$ be another. Then as $$G$$ is connected, there is an edge $$e_1=wu_1$$ incident to $$w$$ such that $$u_1 \in C_1$$, and there is an edge $$e_2=wu_2$$ incident to $$w$$ such that $$u_2 \in C_2$$. Then as every path in $$G$$ from $$u_1$$ to $$u_2$$ contains $$w$$ [because $$u_1$$ and $$u_2$$ are in different components of $$G \setminus \{w\}$$, it follows that there is no cycle containing both $$u_1$$ and $$u_2$$, and thus no cycle containing both $$e_1$$ and $$e_2$$.
The converse is true too. We show the following: If $$G$$ is 2-connected then for every pair $$e_1$$ and $$e_2$$ of edges there is a cycle $$C$$ containing both $$e_1$$ and $$e_2$$. Let us write $$e_1=y_1y'_1$$ and $$e_2=y_2y'_2$$.
1. First let us assume $$y'_2 \not \in \{y_1,y'_1\}$$. Then we claim that there is a cycle $$\hat{C}'$$ in $$G$$ that contains both $$y'_2$$ and $$e_1$$. [Indeed Menger's Theorem says that there is a cycle $$C'$$ that contains $$y_1$$ and $$y'_2$$. If $$C'$$ contains $$y'_1$$ as well then this implies a claimed cycle $$\hat{C}'$$ containing the edge $$e_1=y_1y'_1$$, and $$y'_2$$. [Check this for yourself, draw out $$C'$$ containing $$y_1,y'_1,y'_2$$ and from this draw out the claimed cycle $$\hat{C}'$$.] So assume that $$C'$$ does not contain $$y'_1$$. Then let $$P'$$ be a path from $$y'_1$$ to $$y'_2$$ that does not contain $$y_1$$. Can you put $$C'$$ and $$P'$$ together to get $$\hat{C}'$$? If you aren't seeing this right away, it may be helpful to draw $$C'$$ and then the edge $$e_1=y_1y'_1$$ assuming $$e_1=y_1y'_1 \not \in C'$$, and then to draw $$P'$$ out from $$y'_1$$ to $$y'_2$$, stopping your pencil when $$P'$$ hits a vertex in $$C'$$.]
2. There is a cycle $$C$$ that contains $$e_1=y_1y'_1$$ and $$e_2=y_2y'_2$$. [Indeed, if $$\hat{C}'$$ as in 1. above contains $$y_2$$ as well then we are done. [Indeed, draw $$\hat{C}'$$ out with the edge $$e_1=y_1y'_1 \in \hat{C}$$ and the vertices $$y_2,y'_2 \in \hat{C}'$$ and from this, see if you can find a cycle $$C$$ containing both $$e_1=y_1y'_1$$ and the edge $$e_2=y_2y'_2$$.] Otherwise, $$y_2$$ is not in $$\hat{C}'$$, and so this implies of course that $$y_2 \not \in \{y_1,y'_1\}$$. So now let $$P''$$ be a path from $$y_2$$ to $$y_1$$ that avoids $$y'_2$$. Can you can get the desired cycle $$C$$ containing both edges $$e_1$$ and $$e_2$$ from this? Draw out $$\hat{C}'$$ containing $$e_1=y_1y'_1$$ and $$y'_2$$ but not $$y_2$$, and then the edge $$e_2=y_2y'_2$$, and then draw the path $$P''$$ from $$y_2$$ to $$y_1$$, where $$P"$$ avoids $$y'_2$$, keeping in mind that $$P''$$ may hit another vertex in $$\hat{C}'$$.]
The cycle $$C$$ as in 2. is the desired cycle.