I am seeking advice on how to prove something. Apologies if my terminology is incorrect: I am not a mathematician.
Let $G_1$ and $G_2$ be undirected cycle graphs with edges $E_{G1}$ and $E_{G2}$ respectively.
Let $e_{G1} = (a,b)$ and $e_{G2} = (x,y)$ be arbitrary edges from $E_{G1}$ and $E_{G2}$ respectively.
Remove these edges from their respective graphs.
For every edge in $G_1$, substitute $a$ for $x$ and $b$ for $y$.
Let $G_3 = G_1 \cup G_2$.
$G_3$ is a undirected cycle graph.
I would like to prove this, but I have literally no idea how to do this. Can anyone offer any advice?
Thanks
This is the type of result that's much easier to intuit than it is to rigorously prove. For a rigorous proof, it's probably best to enumerate the vertices and edges of each graph:
Let $G_1$ and $G_2$ be cycles with:
$$V(G_1)=\{u_1, u_2, \dots, u_{m}\}$$
$$E(G_1) = \{u_1 u_2, u_2 u_3, \dots, u_{m-1} u_{m}, u_{m} u_1\}$$
$$V(G_2)=\{v_1, v_2, \dots, v_{n}\}$$
$$E(G_2) = \{v_1 v_2, v_2 v_3, \dots,v_{n-1} v_{n},v_{n} v_1\}$$
You may assume that the edge to be deleted from $G_1$ is $u_m u_1$, and the edge to be deleted from $G_2$ is $v_1 v_n$ because of symmetry (otherwise you could relabel the vertices to make it so). Then the graph $G$ obtained by deleting $u_m u_1$ from $G_1$ and $v_n v_1$ from $G_2$ and identifying vertex $u_1$ with vertex $v_n$ (calling the new vertex $u_1$) and vertex $u_m$ with vertex $v_1$ (calling the new vertex $u_m$) is the graph you are looking for. Now we have:
$$V(G) = \{u_1, u_2, \dots, u_m, v_2, v_3, \dots, v_{n-1}\}$$
$$E(G) = \{u_1 u_2, u_2 u_3, \dots, u_{m-1} u_m, u_m v_2, v_2 v_3, v_3 v_4, \dots, v_{n-2} v_{n-1}, v_{n-1} u_1\}$$
But this is exactly a cycle on $m+n-2$ vertices.