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So... last question that I have to do this semester... and of course it's one that I am completely wedged on.

I am supposed to show that for any discrete subgroup $G$ of the isometry group in the plane, which contains at least one non-trivial translation, there is at least one point $p$ such that if $g,h$ are transformations in $G$, if $g(p)$ = $h(p)$, then $g = h$.

My thoughts so far are that--clearly points brought about by just rotations can be gotten to in different ways by composing different rotations. So it must be some point that involves the translation (and of course that must be why there's at least one translation).

I've also thought about the contrapositive a lot.

But I just can't make anything click. Anyone have some thoughts in the right direction?

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1 Answer 1

Hints:

The problem is equivalent to finding a point which is not fixed by any nontrivial element of G. (Do you see why?)

So take a nontrivial isometry of G. What could its fixed points be? Could the fixed points of all nontrivial elements of G cover the entire plane?

(The statement is true even if $G$ has no nontrivial translations, as long as $G$ has at least one nontrivial element of some kind.)

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