Let $G$ be a connected $7$ regular graph with exactly one cut edge $xy$. Let $G_1$ and $G_2$ be the components of $G-xy$. Prove that if $G_1$ and $G_2$ are $6$-edge connected, then $G$ has a perfect matching.
Here is my approach but got stuck at some point: Suppose for contradiction $G$ has no perfect matching, then By Tutte's Theorem, there exists a set $S \subseteq V(G)$ s.t. number of odd components of $G-S$ $>|S|$. Let $S=\{v_1, v_2, \cdots v_s\}$ and odd component of $G-S$ are $H_1, \cdots, H_t$ where $t \geq s+1$. Then the number of edges between $H_j$ and $S$ is odd for $1 \leq j \leq t$ is odd by the handshaking lemma. Moreover, the number of edges between $H_j$ and $S$ are at least $3$ except for the cut edge $xy$. Therefore, number of edges between $S$ and $V(G)-S$, denoted as $|E(S, V(G)-S)|$, is at least $3(t-1)+1 = 3t-2$, so $|E(S, V(G)-S)| \geq 3t-2$. On the other hand, since $G$ is $7$ regular, $|E(S, V(G)-S)| \leq 7s < 7t$.
But I'm stuck after that to show there is a contradiction about $G_1$ and $G_2$ are $6$edge connectivity. Any help would be appreciated.