abstract algebra - dihedral groups

**dori1123** · September 17th 2008, 05:12 PM

Can someone help me with this proof? It was an exam problem, but I can't get it.

D_2n = < r, s | r^n = s^2 = 1, rs = sr^(-1) >
If n = 2k is even and n>= 4, show that z = r^k is an element of order 2 which commutes with all elements of D_2n. Show also that z is the only nonidentity element of D_2n which commutes with all elemtns of D_2n.

**ThePerfectHacker** · September 17th 2008, 05:30 PM

Originally Posted by dori1123

Can someone help me with this proof? It was an exam problem, but I can't get it.

D_2n = < r, s | r^n = s^2 = 1, rs = sr^(-1) >
If n = 2k is even and n>= 4, show that z = r^k is an element of order 2 which commutes with all elements of D_2n. Show also that z is the only nonidentity element of D_2n which commutes with all elemtns of D_2n.

Hint: Use the fact that $r^n s = r^{-n} s$

**dori1123** · September 17th 2008, 06:57 PM

For the first proof, should I show that r^k is an element of order 2 or should I show that r^k commutes with all elements of D_2n?

Can you help me to get started? Thank you.

**ThePerfectHacker** · September 17th 2008, 07:24 PM

Originally Posted by dori1123

For the first proof, should I show that r^k is an element of order 2 or should I show that r^k commutes with all elements of D_2n?

Note that $r^k \not = e$ -- why?
Now $(r^k)^2 = r^{2k} = e$ .
Thus the order of $r^k$ is two.
This means that $(r^k)(r^k)^{-1} = e$ .
Thus, $r^{-k} = r^k$ .

Any element in the dihedral group has form $r^as^b$ .
Now $(r^as^b)r^k = r^a(s^br^k) = r^a(r^{-k}s^b) = r^a(r^ks^b) = r^k(r^as^b)$ .
Thus $r^k$ commutes with everything.

**dori1123** · September 18th 2008, 07:37 AM

Originally Posted by ThePerfectHacker

Note that $r^k \not = e$ -- why?
Now $(r^k)^2 = r^{2k} = e$ .
Thus the order of $r^k$ is two.

So I know that $r^k \not = e$ and $(r^k)^2 = r^{2k} = e$ by the definition of $D_{2n}$ , because $n = 2k$ , so $r^n = r^{2k} = (r^k)^2 = e$ ?

**ThePerfectHacker** · September 18th 2008, 07:53 AM

Originally Posted by dori1123

So I know that $r^k \not = e$ and $(r^k)^2 = r^{2k} = e$ by the definition of $D_{2n}$ , because $n = 2k$ , so $r^n = r^{2k} = (r^k)^2 = e$ ?

Yes

**dori1123** · September 18th 2008, 12:49 PM

How do I show that z is the only nonidentity element of $D_{2n}$ which commutes with all elements of $D_{2n}$ ?

**ThePerfectHacker** · September 18th 2008, 12:54 PM

Originally Posted by dori1123

How do I show that z is the only nonidentity element of $D_{2n}$ which commutes with all elements of $D_{2n}$ ?

Hint: The element which commutes with all element if and only if it commutes with $r,s$ . Thus find which element commutes with $r,s$ .

**dori1123** · September 19th 2008, 05:55 AM

Obviously, the identity element commutes with all elements in $D_{2n}$ . But I am looking for a nonidentity element, so from the first proof, $r^k$ is the nonidentity element that commutes with all elements in $D_{2n}$ . But how do I know that it is the ONLY nonidentity that commutes with those elements?

**ThePerfectHacker** · September 19th 2008, 09:40 AM

Originally Posted by dori1123

But how do I know that it is the ONLY nonidentity that commutes with those elements?

Look at what I wrote above! Let $x$ be an element in this group then we can write it as $x=r^as^b$ where $0\leq a\leq n-1, 0\leq b\leq 1$ . Now $r^as^b$ commute with all the element if and only if it it commutes with the generating set i.e. it commute with $r$ and $s$ . Thus, find the conditions on $a,b$ so that $r^as^b$ commute with $r$ and $s$ .

**dori1123** · September 20th 2008, 06:43 PM

So I let $x = r^as^b$ where $0<=a<=n-1, 0<=b<=1$ , then $(r^as^b)r = r^a(s^br) = r^a(r^{-1}s^b) = r^ars^b = r(r^as^b)$ and $(r^as^b)s = (r^as)s^b = sr^{-a}s^b = s(r^as^b)$ . So $r^as^b$ commutes with all elements of $D_{2n}$ no matter what a and b are, isn't it? Then the nonidentity element in the form of $r^as^b$ also commutes with all elements of $D_{2n}$ , so $r^a$ is not the only nonidentity element that commutes with the elements in $D_{2n}$ ...?
Am I thinking this right?...

**ThePerfectHacker** · September 20th 2008, 07:52 PM

Originally Posted by dori1123

So I let $x = r^as^b$ where $0<=a<=n-1, 0<=b<=1$ , then $(r^as^b)r = r^a(s^br) = r^a(r^{-1}s^b) = r^ars^b = r(r^as^b)$ and $(r^as^b)s = (r^as)s^b = sr^{-a}s^b = s(r^as^b)$ . So $r^as^b$ commutes with all elements of $D_{2n}$ no matter what a and b are, isn't it? Then the nonidentity element in the form of $r^as^b$ also commutes with all elements of $D_{2n}$ , so $r^a$ is not the only nonidentity element that commutes with the elements in $D_{2n}$ ...?
Am I thinking this right?...

Let us consider two cases: (i) $x=r^a$ , $0\leq a\leq n-1$ (ii) $x=r^as$ , $0\leq s\leq 1$ .

In case (i) we have that $x=r^a$ commutes with $s,r$ . It definitely commutes with $r$ no matter what $a$ is but to commute with $s$ we need to have $r^a s = sr^a = r^{-a}s$ and so $a\equiv -a ~ (n)$ this forces $a = k$ (where $k$ is such an integer so $2k=n$ ). Thus, we find that $x=r^k$ commutes with all the elements.

You try case (ii).

**dori1123** · September 20th 2008, 08:32 PM

So, $(r^as)s = (sr^{-a})s = (sr^a)s = s(r^as)$ . So $r^as$ commutes with s. And $(r^as)r = (sr^{-a})r = (sr^a)r$ . Here is what I wonder, if I am given $r^as = sr^{-a}$ , does that mean $sr^a = r^{-a}s$ ?

**ThePerfectHacker** · September 21st 2008, 03:50 PM

Originally Posted by dori1123

So, $(r^as)s = (sr^{-a})s = (sr^a)s = s(r^as)$ . So $r^as$ commutes with s. And $(r^as)r = (sr^{-a})r = (sr^a)r$ . Here is what I wonder, if I am given $r^as = sr^{-a}$ , does that mean $sr^a = r^{-a}s$ ?

$(r^a s)s = r^a s^2 = r^a$ .
$s(r^a s) = (sr^a)s = (r^{-a} s)s = r^{-a}$ .
This means if we want commutavity then $r^a = r^{-a}$ i.e. $a=0,k$ .

Now there are two cases: (i) $x=r^0 s = s$ (ii) $x=r^k s$ .
Which of these commute with the remaining element $r$ ?

**dori1123** · September 21st 2008, 04:49 PM

Originally Posted by ThePerfectHacker

Now there are two cases: (i) $x=r^0 s = s$ (ii) $x=r^k s$ .
Which of these commute with the remaining element $r$ ?

So for $x = r^0s = s$ , $xr = sr$ does not equal to $rx = rs = sr^{-1}$ (unless $r = r^{-1}$ ?). For $x = r^ks$ , $xr = r^ksr = sr^{-k}r = sr^{-k+1}$ and $rx = rr^ks = r^{k+1}s$ , so $r^ksr$ does not equal to $rr^ks$ . Right?

Thread: abstract algebra - dihedral groups

LinkBack

Thread Tools

Display

abstract algebra - dihedral groups

Similar Math Help Forum Discussions

Abstract Algebra-Groups

Abstract Algebra: Groups

Abstract Algebra: Groups

Abstract Algebra: Groups

Abstract Algebra: Groups and Symmetry

Search Tags