Look at this page:
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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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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