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Continuing my independent journey through "Abstract Algebra" (see this previous question for context and notation), I am working on:

If $n = 2k$ is even and $n \geq 4$, show that (a) $z = r^k$ is an element of order 2 which (b) commutes with all elements of $D_{2n}$ (Dihedral group of a regular n-gon). Show also that (c) $z$ is the only non-identity element of $D_{2n}$ which commutes with all elements of $D_{2n}$.

(a) $z^2 = (r^k)^2 = r^{2k} = r^n = 1$, so the order of $z$ divides 2.
$\quad\quad$ Does the hypothesis that $n \geq 4$ come into play here?

(b) Elements in $D_{2n}$ are of the form $r^i$ or $sr^i$, $0 \leq i \leq n - 1 $

$\quad\quad$ For elements $r^i$, we have: $zr^i = r^kr^i = r^{k + i} = r^{i + k} = r^ir^k = r^iz$,
$\quad\quad$ so $z$ commutes with these elements.

$\quad\quad$ For elements $sr^i$, I am a little confused (yet still trying to remember what needs to be shown!):

$\quad\quad$ $zsr^i = r^ksr^i = (r^ks)r^i = (sr^{-k})r^i = sr^{-k + i}$
$\quad\quad$ Yet $sr^iz = sr^ir^k = sr^{k + i}$
$\quad\quad$ The two don't seem equal to me, so I must be missing something?

(c) The only thing that comes to mind is the argument that the identity in a group is unique,
$\quad\quad$ but I'm having a hard time generalizing/extending that concept to this group.

Thanks in advance for any hints!

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  • $\begingroup$ By the way: "commutator" has a special meaning: a commutator in a group $G$ is an element of the form $xyx^{-1}y^{-1}$, for $x,y\in G$ (rather than an element that "commutes" with other elements). So I don't think your title says what you wanted to say. $\endgroup$ May 18, 2011 at 21:13
  • $\begingroup$ @Arturo: I suppose the title isn't what I intended! What is the correct term? $\endgroup$
    – Altar Ego
    May 19, 2011 at 7:23
  • $\begingroup$ An element that commutes with everything is a "central element"; when talking about things commuting, "centralizing" (or "centralising") and its variants are common terms. Here, you are really looking at central elements in the dihedral group (the order is also not $n/2$: the order of the element in question is $2$. It's the $(n/2)$th power of the rotation, but the order of the element is $2$. $\endgroup$ May 19, 2011 at 16:05
  • $\begingroup$ @Arturo: Got it. (n/2)*2 = n :) $\endgroup$
    – Altar Ego
    May 19, 2011 at 17:04

3 Answers 3

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For $(a)$, if $n$ is even and less than $4$, then $n = 2 $ (and you have the non-cyclic group of order $4$, the symmetry group of the slightly bizarre "digon") or $n$ is not positive, and there is no dihedral group. Some people don't even like calling the dihedral group of order $ 4$ a dihedral group. The conclusion of $(a) $ and $(b)$ are true when $n=2$, but $(c)$ is not.

For $(b)$, I assume you are worried about claiming $r^k = r^{-k}$, as the rest of the expression is the same (and so could be canceled). They certainly look different, but if consider the special relation k and n have, I think you'll see they are the same.

For $(c)$, I admit I would use the description of dihedral groups of order 2n as consisting of n rotations and n flips. "Adjacent" flips (for $n\ge3$) do not commute, so no flip is in the center. If a rotation is in the center, then it must be the same as its flip. Only one rotation $r^K$ could possibly satisfy $r^K = r^{-K}$, and this is the key idea of $(b)$ again: $K=k $ has a very special relationship to n.


A conceptual thing for later: when you look at "normalizers" and "centralizers" one really bizarre and useful thing is that the normalizer of a group of order $2 $is always equal to its centralizer. This is why $z$ is in the center: $z^*z = 1$, and $z$ is normalized by the flips.

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(a) The fact that the order is exactly $2$ (and not merely divides $2$) now follows simply because the order of $r$ is exactly $n$, and $0\lt k\lt n$, so $z=r^k\neq 1$.

No, the hypothesis that $n\geq $ does not affect the fact that $z$ is of order $2$. This holds in all dihedral groups. That hypothesis has to do with (c) (for example, if you take $D_4$, then every element commutes with every other element, as you should have little trouble verifying, so that the assertion in (c) would be false then).

(b) Remember part (a)! You proved that $z^2 = e$, so $z=z^{-1}$. That means that $r^k = (r^k)^{-1} = r^{-k}$. You have $z(sr^i) = sr^{-k+i} = sr^{-k}r^i$; but since $r^{-k}=r^k$, then...

(c) You have already shown that every element commutes with $r^k$. Now you need to show that if $s^jr^i$, $0\leq j\leq 1$, $0\leq i\leq n-1$ commutes with everything, then either $s^jr^i = r^k$, or $s^jr^i = 1$.

Show that if $0\lt i\lt n$, and $r^i$ commutes with $s$, then $i=k$. That will show that the only power of $r$ that commutes with everything is $r^k=z$ (this is where you will need that $n\geq 4$).

Now think about elements of the form $sr^i$. If they commute with everything, then they must commute with both $s$ and with $r$. So, for example, you must have that $s(sr^i) = s^2r^i = r^i$ is equal to $(sr^i)s = s^2r^{-i} = r^{-i}$; when do you have $r^i = r^{-i}$? So that whittles down the possible elements that can commute with everything. And now, what happens when you compute $r(sr^i)$ and $(sr^i)r$? When, if ever, do you get equality?

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Used the formula

If order$(r)= n $, then $r^is=sr^{-i}$ for $0 \le i \le n-1$

$D_{2n} = \{1,r,r^2,.....,r^{n-1},s,sr,....,sr^{n-1}\}$

Now we have to prove this $3$ result given below

$1)$ $r^k$ has order $2$

we can easily see that $(r^k)^2= r^{2k} =r^n=1$ ,

since $|r|=n$ then it implies $r^{k}\neq 1$

$2)$ $r^k$ commutes with all element in $D_{2n}$

i,e $(r^k)(r^i)=r^{k+i} = (r^i)(r^k)$ for all $i=0,1,2,...,n-1$ and for $0 \le i \le n-1$,

Now from the given above formula, we have $r^k=sr^{-k} , r^{2k}=1$ implies $r^k=r^{-k} $ and $r^kr^i=r^ir^k$

Therefore $(r^k)(sr^i)=(r^ks)(r^i)=(sr^{-k})(r^i)=(sr^k)(r^i)=(sr^i)(r^k)$

So, $r^k$ commutes with all of the $sr^i$ ,this showed that $ r^k$ commutes with every element in $D_{2n}$

$3)$ suppose $x \in D_{2n}$ and $x \neq 1$ .

If $ x=r^i$, for all $1 \le i \le n-1 $ and $i \ne k$ then $ x$ doesnot commute with $s$

because if $ x$ commute with $s$, then $r^is=sr^i=r^{-i}s$ this implies that $r^i=r^{-i}$

i,e $r^{2i}=1$. Now we have $2i \ge n$ since order$(r)=n$

$2i \ge n$ implies $ 2i-n \ge 0$ so $ r^{2i-n}=1$ but we know that $1\le i\le n-1$ after multiply by $n$ we get $n \le 2i \le 2n-2$ implies $0 \le 2i-n \le n-2$

according to the given question we have $n=2k$ i,e $k= \frac{n}{2}$

Now come again in the line , we have $0 \le 2i-n \le n-2$ this $2i=n$ this will give $i= \frac{n}{2}$ . lastly we are getting contradiction because we have already assume that $i\ne k$ in $(3)$

Now we have to show that $r^k$ is the only non identity element that commutes with all element in $D_{2n}$

Proof : from the given above in $(3)$, we have already showed that if $ x=r^i$, for all $1 \le i \le n-1 $ and $i \ne k$ then $ x$ doesnot commute with $s$

let us assume again it commutes i,e

$r(sr^i)=(sr^i)r $

$sr^{-1}r^i=sr^ir $

$r^{i-1}=r^{i+1}$

$r^2=1$

But according to the given question we have $n \ge 4$, so we have contradiction here

therefore ,$r^k$ is the only non idenity element that commutes with all elements in $D_{2n}$

also $z$ is the only non idenity element that commutes with all elements in $D_{2n}$

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