# A quotient manifold with injectivity radius $0$

Let $$M$$ be a Riemannian manifold equipped with a free and proper isometric action by a Lie group $$G$$. Suppose that $$M$$ has positive injectivity radius.

Question: Is it possible for the quotient manifold $$M/G$$, equipped with the quotient Riemannian metric, to have injectivity radius $$0$$?

Thoughts: For this to happen, the action must be such that the quotient $$M/G$$ is non-compact. I think one way to construct an example would involve a $$G$$-action where the $$G$$-orbits in $$M$$ "shrink in size" as one approaches a particular point of $$M$$, similar to the rotational action of $$S^1$$ on $$\mathbb{R}^2$$ (except this is not free). I suspect there is a simple example, but can't quite see it.

The simplest example is when $$(M, g)$$ is the upper half plane $$\mathbb H$$ with the hyperbolic metric
$$g = \frac{1}{y^2} (dx^2+ dy^2),$$
while $$G = \mathbb Z$$ is the action of the translation
$$(x, y) \mapsto (x+n, y).$$