A connected k-regular bipartite graph is 2-connected.

I’ve been struggling with this exercise; all ideas have been unfruitful, leading to dead ends. It is from Balakrishnan’s A Textbook of Graph Theory, in the connectivity chapter:

Prove that a connected k-regular bipartite graph is 2-connected.

(That is, deletion of one vertex alone is not enough to disconnect the graph).

I think the objective is to make use of Whitney’s theorem according to which a graph (with at least 3 vertices) is 2-connected iff any two of its vertices are connected by at least two internally disjoint paths. But I’ll welcome any ideas or solutions.

Thank you!

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Let $G = (V_{1}\cup V_{2},E)$ be a connected, $k$-regular bipartite graph where $V_{1}$ and $V_{2}$ are the partite vertex sets. As the case $k=1$ is trivial, we may assume that $k \geq 2$ and therefore $|V_{1}\cup V_{2}|\geq 4$.

Assume for contradiction that $G$ is not $2$-connected.

As $G$ is connected but not $2$-connected, there exists a vertex $v$ whose removal disconnects the graph. Without loss of generality we may assume that $v \in V_{1}$. Then $G-v = \uplus_{i\in [1,a]} G_{i}$ where each $G_{i}$ is a connected component and $a \geq 2$.

As $a \geq 2$, there exists some component $G_{b}$ such that $|V_{1}\cap V(G_{b})| \geq |V_{2}\cap V(G_{b})|$. (It shouldn’t be too hard to convince yourself of this) For convenience denote $L = V_{1}\cap V(G_{b})$ and $R = V_{2}\cap V(G_{b})$.

As $G_{b}$ is a connected component and $G$ was connected, and $v \in V_{1}$, at least one vertex in $R$ was adjacent to $v$, and therefore has degree less than $k$. However the vertices in $L$ have lost no edges Then we have
$$
\sum_{u\in R}deg(u) < k\cdot|R| < k\cdot|L| = \sum_{w \in L}deg(w)
$$
However as $G[L \cup R]$ forms a bipartite graph, we know
$$
\sum_{u\in R}deg(u) = \sum_{w \in L}deg(w)
$$

Thus we have a contradiction, so $G$ must be at least $2$-connected.

Proof by contradiction. Suppose $v$ is a cut vertex of $G = G(X,Y).$ Without loss of generality, let $v\in Y.$ Then $G-v$ has at least two components, say $C_1$ and $C_2.$ Let $(X_1,X_2)$ be the bipartition of $C_1.$ Let $s$ be the number of neighbors of the vertex $v$ in $C_1$ (that is, in $X_1$), then $s<r.$

Since $C_1$ is bipartite, the number of edges incident at $Y_1$ in $C_1$ must be equal to the number of edges incident at $X_1$ in $C_1.$ Hence
$$ r|Y_1| = s(r-1) + (|X_1| – s)r$$
$$\Rightarrow s = (|X_1| – |Y_1|)r$$
$$\Rightarrow s \geq r,$$
a contradiction.