2-edge connected has perfect matching, then graph has two perfect matching Let $G$ be a graph with at least 1-factor. Prove that if $G$ is 2-edge-connected, then $G$ has at least two perfect matchings.
i tried it by Tutte's theorem first,
I choose $e$(edge) in first perfect matching, 
and when G doesn't have two perfect matching, then there exist $X$ in $G$ ,$odd(G-e-X)>|X|$ 
by pareity, $odd(G-X)=|X|$ 
and i was blocked here. how can i solved it?
 A: Prove it by induction on the number of vertices in $G$. The base case is graph with 4 vertices. It is easy to show that a 2-connected graph on 4 vertices contains a 4 cycle, so it contains at least 2 perfect matchings. 
Now, consider a 2-connected graph $G$ with at least 4 vertices and let $M$ be a perfect matching of $G$. Consider a walk starting from a vertex $v$ that alternately takes an edge in $M$ and not in $M$. Note that such a walk exists since $M$ is a perfect matching and each vertex has degree at least 2). At some point the walk will visit a vertex for the second time, hence closing a cycle $C$. If $C$ is an even cycle, then it is an alternating cycle and $C\triangle M$ is another matching in $G$. If $C$ is odd, we can apply induction on $G/C$, since $G/C$ is 2-connected and $M/C$ is a perfect matching of $G/C$. Combine the other perfect matching $M'$ in $G/C$ with a matching in $C$ that covers every vertex expect for the single vertex in $C$ covered by $M'$ to get the other perfect matching in $G$.
