# prove that a graph is one-ended

I am not sure how to show the second part. Isn’t it obvious from the fact that G and H are infinite?

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The following was a bit too long for a comment so I wrote it as an answer:

In my opinion it is better to look at this problem in a pure structure theoretical fashion, since your assertion is true for any pair of connected infinite locally finite graphs. So nothing special about the graphs being Cayley graphs of some groups.

Hence: Let $X$ and $Y$ be two connected infinite locally finite graphs. Then their product $X \times Y$ has exactly one end.

There are lots of “different products” one could form with two given graphs. In this case we have $V(X \times Y) = V(X) \times V(Y)$ and the adjacency is defined as follows:

$(x_1, y_1)$ is adjacent to $(x_2, y_2)$ iff $x_1 = x_2$ and $y_1$ is adjacent to $y_2$ or vice versa.

Let now $B$ be some ball of finite radius in $X \times Y$ and let $z,w$ be two vertices outside of $B$ and each inside some infinite component. If you find a path connecting them outside of $B$ you are done.

First think about the example $\mathbb{Z} \times \mathbb{Z}$ and some ball in its Cayley graph (w.r.t some generating set). Why can you connect any two vertices outside this ball without touching the ball (consider the definition of the adjacency in such product graphs at the same time)?

I am confident that you can write it down in general.