# Group theory, order with permutation of Z

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.

• Fix all but a finite number of elements. This gives you an element of finite order, as you'd expect. – Alex Wertheim Feb 17 '15 at 3:30
• A non-trivial example of an element of finite order is the one that switches $2n$ with $2n+1$ for all integers $n$. – Stephen Montgomery-Smith Feb 17 '15 at 3:32
• @AlexWertheim If you take two of those, however, their product aso fixes all but finitely many, so the product does not have infinite order. – Thomas Andrews Feb 17 '15 at 3:41
• @ThomasAndrews: yes, indeed. I was only trying to give an example of an element of finite order, not a solution. – Alex Wertheim Feb 17 '15 at 4:06

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.
• 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 $f_1f_0$. – pjs36 Feb 17 '15 at 19:52
• $f^2_0$ would just go back to the original $x$ value? – All About Groups Feb 17 '15 at 19:56
• Yep, as would $f_1^2$, meaning you've found two elements of finite order. – pjs36 Feb 17 '15 at 20:08
• 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 $x$. – pjs36 Feb 17 '15 at 20:14