I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!


Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:

(a) Z(D$_{2n}$) = 1 if $n$ is odd.

(b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.

Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }


part (a)

Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.

For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.

Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.

Then we have s$^j$r$^w$$r$ = $r$s$^j$r$^w$. Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$

In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.

Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.

Then $s$$r$$^w$ = $r$$^w$$s$. Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.

Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:

$$ \bar{w} = (n-1)*w $$

where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$

$$ w + an = wn - w $$ $$ 2w = wn - an $$ $$ 2w = n(w-a) $$

Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.

Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.

part (b)

From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.

$$ 2w = n(w - a) $$

Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired.

  • 2
    $\begingroup$ You could remind us what $r$ and $s$ denote. Also, using mathjax consistently throughout would make the question more readable. $\endgroup$ May 25, 2019 at 3:40
  • $\begingroup$ @LordSharktheUnknown, ive updated the post with the group presentation using r and s as generators and their associated relations $\endgroup$
    – H_1317
    May 25, 2019 at 4:00
  • $\begingroup$ Thanks, but again your mathjax is poor. It is best to put all mathematical formulae into mathjax, including single letters used as variables. Also, each complete formula should be enclosed in dollars, not the single letters/symbols within it. This ensures that the formula is spaced well, and involves a lot less effort on the writer's part. $\endgroup$ May 25, 2019 at 4:04
  • $\begingroup$ And $r$ is still introduced (in the statement of the problem) before it is defined some lines later. $\endgroup$ May 25, 2019 at 5:40

1 Answer 1


$D_{2n}$ has presentation $\langle r,f\mid r^n,f^2,(rf)^2\rangle $.

If $n$ is odd, $f,r,r^2,\dots, r^{n-1}$ are not in the center, since $rf=fr^{-1}$.

Similarly, $r^af,1\le a\le n-1$ is not, because $(r^af)f=r^a$ but $f(r^af)=ffr^{-a}=r^{-a}$.

That leaves just $\{e\}$ for $n$ odd.

We get $\{e,r^{\frac n2}\}$, for $n$ even, since $r^{\frac n2}f=fr^{\frac n2}$.

  • $\begingroup$ Thanks, but i was more hoping for a review of my proof. Is my proof correct? $\endgroup$
    – H_1317
    Jun 3, 2019 at 3:04
  • $\begingroup$ Sure. I'm working on it, but finding it a little hard to follow. It may just be me. I understand about $\dfrac 23$ of it. $\endgroup$
    – user403337
    Jun 3, 2019 at 3:08
  • $\begingroup$ If theres a particular part u find confusing, i will clean that part up for you $\endgroup$
    – H_1317
    Jun 3, 2019 at 3:13
  • $\begingroup$ For instance, in part $(a)$, when you say we have arrived at the task of determining when $w=-w$ as powers of $r$... could certainly be worded better. That is, maybe write it mathematically. $\endgroup$
    – user403337
    Jun 3, 2019 at 3:19
  • 1
    $\begingroup$ I'm getting the impression that it's pretty correct. You did go into a lot of detail. I commend you. I like the changes you made, too. $\endgroup$
    – user403337
    Jun 3, 2019 at 3:56

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