Proof of Uniqueness of Non-Identity Commuting Element in D_2n

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Homework Statement


If ##n = 2k## is even and ##n \ge 4##, show that ##z = r^k## is an element of order 2 which commutes with all elements of ##D_{2n}##. Show also that ##z## is the only nonidentity element of ##D_{2n}## which commutes with all elements of ##D_{2n}##.


Homework Equations


The question also says to use this previously-proven result: If ##x## is an element of finite order ##n## in a group ##G, n = 2k##, and ##1 \le i < n##, then ##x^i = x^{-i}## if and only if ##i = k.##


The Attempt at a Solution


I have managed to show everything except the uniqueness of such an element (which is normally the easy part!). I know that I need to assume that another such element exists, and use this assumption to show that this element is in fact ##z##, giving a contradiction. However, I am struggling to generate such a contradiction; I always seem to end up with trivial equations of the type ##1 = 1## (where ##1## is the identity). I think I just need a quick hint as to how to generate my contradiction.

Thanks!
 
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Take an element ##r^a s^b## in ##D_{2n}##. Thus, we have ##1\leq a\leq n## and ##b = 0,1## that commutes with every element.

So take an arbitrary element ##r^x s^y##. Then we have

r^a s^b r^x s^y = r^x s^y r^a s^b

Try to write both sides of the above equation in the form ##r^q s^p##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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