Prove that a 3-regular graph $G$ has a cut-vertex if and only if $G$ has a bridge.

Question

Prove that a 3-regular graph $G$ has a cut-vertex if and only if $G$ has a bridge.

Here is what I got so far

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Assume that $G$ is a 3-regular graph and $G$ has a bridge. Let $u,v \in V(G)$ such that $uv$ is a bridge in $G$. Since $uv$ is a bridge, either $U$ or $V$ or both must be cut-vertex. So $G$ contains a cut-vertex.

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Assume that $G$ is a 3-regular graph and $G$ has a cut-vertex. Show that $G$ has a bridge. I tried to find a way to use the fact that $G$ is 3-regular graph, to show $G$ has a bridge, but I haven't success so far.

Answer 1

Let $v$ be a cut-vertex of $G$. Removing $v$ from $G$ create 2 or 3 connected components, since $v$ has 3 neighbors.
Furthermore, one of these components must contain exactly one neighbor of $v$ (if it's 2 components, one has two and the other has one. If it's 3 components, all have one).
Let $Y$ be the vertices of such a component containing exactly one neighbor $v'$ of $v$.

We get that for any $y \in Y$ other than $v'$, the path from $y$ to $v$ passes by $v'$, as $v'$ is the sole neighbor of $v$ in $Y$. Therefore, removing $v'$ from $G$ disconnects $y$ from $v$, and $v'$ is a cut-vertex. And since both $v$ and $v'$ are cut-vertices, they form a bridge. Note that $y$ has to exist, something you can probably show on your own :)

Answer 2

Conversely, suppose that e=uv is a cut edge of $G$. Then the deletion of a results is the deletion of the edge uv. Since $G$ is cubic, G-u is disconnected. Hence u is a cut edge of $G$.