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What is the “$q$” in the answer? And what "$0\leq a_0$" means? It's a typo. And I don’t understand the last 2 sentences: If $q≥2$ we have $2a<2n ≤ qn$, a contradiction. If $q=1$, we have $n=2n$ even, a contradiction.

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Under the hypothesis that $x$ is a power of $r$, then $a$ must be $0\leq a\leq n-1$. – Sigur Jan 23 '13 at 0:49
That page is pretty bad. I wouldn't worry about it. – Gerry Myerson Jan 23 '13 at 0:58
who can help me understand what the proof in that page means? – i_a_n Jan 23 '13 at 4:20
Alexander has supplied a better proof. If you are not familiar with Alexander's notation $[x,y]$, it just means $x^{-1}y^{-1}xy$, and it yields the identity element if and only if $x$ commutes with $y$. – Gerry Myerson Jan 24 '13 at 0:04

1 Answer 1

I don't think that proof is very good, so I will explain the general case (for arbitrary $n$). Then if you like you can go back and figure out what the link is doing.

First, the presentation of $D_{2n}$ provides $r^s=r^{-1}$. Thus we have $[r,s]=r^{-2}=r^{n-2}$, which, since $rs=sr[r,s]$, tells us that any given word in $r$ and $s$ can be written in the form $s^ar^b$, where $a=0$ or $1$ and $b=0,\ldots,n-1$. (Why?) Thus any $g\in D_{2n}$ has the form $g$ has the form $g=s^ar^b$.

You want to find which elements $z=s^{x}r^{y}$ so that $[z,g]=1$ for all $g\in D_{2n}$. Let's calculate: $$ \begin{eqnarray*} [z,g]&=&z^{-1}g^{-1}zg\\ &=&(s^xr^y)^{-1}(s^ar^b)^{-1}(s^xr^y)(s^ar^b)\\ &=&r^{-y}s^{-x}r^{-b}s^{-a}s^xr^ys^ar^b\\ &=&r^{-y}s^xr^{-b}s^as^xr^ys^ar^b\\ \end{eqnarray*} $$ Note here that if $x=0$, $$ \begin{eqnarray*} &=&r^{-y}r^{-b}s^ar^ys^ar^b\\ \end{eqnarray*} $$ Obviously this is $1$ when $a=0$. When $a\not=0$, $$ \begin{eqnarray*} &=&r^{-y}r^{-b}sr^ysr^b\\ &=&r^{-y}r^{-b}r^{-y}r^b\\ &=&r^{-y-b-y+b}\\ &=&r^{-2y}\\ \end{eqnarray*} $$ which is $1$ only when $y=0$ or $n$ is even and $y=n/2$. A similar calculation for the $x=1$ case (which you should do) yields an expression which cannot be equal to $1$ for an arbitrary $x$ and $y$. Thus the central elements of $D_{2n}$ are $1$ and $r^{n/2}$ if $n$ is even, or only $1$ if $n$ is odd.

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What does $r^s = r^{-1}$ mean in the first line of the proof? I'm not sure what $r^s$ would mean here. – Rori Sep 30 at 13:43
@Rori $r^s\triangleq s^{-1}rs$. It's a group theory shorthand for conjugation. – Alexander Gruber Oct 2 at 2:59
Ah, thanks. I haven't seen that notation. – Rori Oct 2 at 3:03

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