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Let $S_{Z}$ be the set of permutations of ${Z}$ (note that this is an infinite group!). Find two elements of $S_Z$ which both have finite order, but whose product has infinite order.

I just am really struggling even picturing what an element of finite order would look like in this group.

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  • $\begingroup$ Fix all but a finite number of elements. This gives you an element of finite order, as you'd expect. $\endgroup$ – Alex Wertheim Feb 17 '15 at 3:30
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    $\begingroup$ A non-trivial example of an element of finite order is the one that switches $2n$ with $2n+1$ for all integers $n$. $\endgroup$ – Stephen Montgomery-Smith Feb 17 '15 at 3:32
  • $\begingroup$ @AlexWertheim If you take two of those, however, their product aso fixes all but finitely many, so the product does not have infinite order. $\endgroup$ – Thomas Andrews Feb 17 '15 at 3:41
  • $\begingroup$ @ThomasAndrews: yes, indeed. I was only trying to give an example of an element of finite order, not a solution. $\endgroup$ – Alex Wertheim Feb 17 '15 at 4:06
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Thinking geometrically, reflections are nice elements of finite order. So, the map $$x \mapsto -x$$ (which reflects about 0), or $$x \mapsto -(x-1)+1$$ which moves everything 1 unit to the left, then reflects across 0, then moves everything back 1 unit to the right; this accomplishes a reflection across 1.

It turns out that, generally, a product of reflections is a rotation, I wonder if it's some kind of unusual rotation here...

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  • $\begingroup$ So one of the elements will be -1, and then I need another element that has finite order is this correct? $\endgroup$ – All About Groups Feb 17 '15 at 19:32
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    $\begingroup$ Call the first map $f_0$ (since it 'flips' the real line about 0), call the second map $f_1$ (since it 'flips' the real line about 1). Compute $f_0^2 = f_0 \circ f_0$, as well as $f_1^2$. Then compute $f_1f_0$. $\endgroup$ – pjs36 Feb 17 '15 at 19:52
  • $\begingroup$ $f^2_0$ would just go back to the original $x$ value? $\endgroup$ – All About Groups Feb 17 '15 at 19:56
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    $\begingroup$ Yep, as would $f_1^2$, meaning you've found two elements of finite order. $\endgroup$ – pjs36 Feb 17 '15 at 20:08
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    $\begingroup$ Exactly, the composition $f_1f_0 : x \mapsto x + 2$ will have infinite order, since applying it $n$ times will only send $x$ to $x + 2n$, never to simply $x$. $\endgroup$ – pjs36 Feb 17 '15 at 20:14

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