# Proof of Uniqueness of Non-Identity Commuting Element in D_2n

1. Jul 14, 2013

### esorey

1. The problem statement, all variables and given/known data
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 $D_{2n}$.

2. Relevant 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 $i = k.$

3. 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!

Last edited: Jul 14, 2013
2. Jul 14, 2013

### micromass

Staff Emeritus
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 $r^q s^p$.

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