this problem is on dummit, "abstract Algebra",

it says

Let $\displaystyle n\in N$, with $\displaystyle n\geq3$. Prove the following:

(a). Z($\displaystyle D_{2n})=1$ if n is odd.

(b). Z($\displaystyle D_{2n})={ 1,r^{k} }$ if n=2k.

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- Sep 28th 2009, 03:42 PM #1

- Joined
- Feb 2009
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- Sep 29th 2009, 10:54 AM #2
According to my word processor I've spent the last 52 minutes trying to solve this by induction. Which is silly.

I gave a short bit of thought before this, wondering if there was an easier, more subtle proof. I have just realised that there is. So here is a definition and a hint:

We can define the group thus: $\displaystyle D_{2n} = <\alpha, \beta | \alpha^n=\beta=1, \beta \alpha \beta = \alpha^{-1}>$.

Now, when does the equality $\displaystyle a \equiv -a \text{ mod } n$ hold, and why is this relevant here?