Why graph $G = (V, E)$ is elementary cycle? If graph $G = (V, E)$ is bi-connected, and $\forall $ pair $(u, v)$ $u, v \in V$, $(u, v) \notin E$ graph $G - u - v$ is disconnected $\Rightarrow$ graph $G$ is the elementary cycle. 
 A: I'm going to assume you're trying to prove:
Theorem:  Let $G=(V,E)$ be a simple biconnected graph on $n=|V|$ vertices where $n \geq 3$.  If $G$ satisfies the property $P$: "for all distinct $u,v \in V$, if $uv \not\in E$ then $G \setminus \{u,v\}$ is disconnected", then $G=C_n$ or $G=K_n$.
Proof sketch (since this is tagged homework):
Base case:  The theorem is true for $n=3$, by inspection.
Inductive step:  Assume the theorem is true for $m$-vertex graphs, for $m \in \{3,4,\ldots,n-1\}$.


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*If $G$ satisfies $P$ and $G \neq K_n$, then $G \setminus \{u,v\}$ is disconnected for some $u,v \in V$.  Call the disconnected components $H_1,H_2,\ldots,H_t$.  [[Need to show that $t=2$.]]

*Let $M_1$ and $M_2$ respectively be the subgraphs of $G$ induced by the vertices $V(H_1) \cup \{u,v\}$ and $V(H_2) \cup \{u,v\}$, together with the edge $uv$.  Use the inductive hypothesis on $M_1$ and $M_2$ to show that $G=C_n$.  [[Need to show (a) that both $M_1$ and $M_2$ have strictly less than $n$ vertices, (b) that $M_1$ and $M_2$ are also biconnected and satisfy $P$, (c) that, if $M_1$ or $M_2$ were complete graphs on $4$ or more vertices, then a contradiction arises, and (d) that $G=M_1 \cup M_2 \setminus \{uv\}$ is $C_n$.]]
