Reflection and Rotation Centers in D_n: Finding C(F) and C(R) for Even and Odd n

[fishturtle1]

Homework Statement

Let $F$ be a reflection in the dihedral group $D_n$ and $R$ be a rotation in $D_n$. Determine $C(F)$ when $n$ is odd. Determine $C(F)$ when $n$ is even. Determine $C(R)$.

Relevant Equations

Let $a$ be an element of a group $G$. Then $C(a) = \lbrace x \in G : ax = xa \rbrace$

For a reflection $F$ and a rotation $R$ in a dihedral group $D_n$, we have $FR^kF = R^{-k}$ for any integer $k$.

Let $F_0$ be a reflection in $D_n$ s.t. $F \neq F_0$. Observe, $F_0F = FF_0$ is equivalent to $F_0FF_0F = (F_0F)^2 = R_0$. Since a reflection followed by a reflection is a rotation, and the only rotation of order 2 is $R_{180}$, we have $F_0F = R_{180}$. Thus, $F_0F = FF_0$ is equivalent to $F_0F = R_{180}$. Thus, the only reflection $C(F)$ can contain(besides $F$) is $F_0$ such that $FF_0 = R_{180}$.

Next, let $R_x$ be any rotation in $D_n$. Observe, $FR_x = R_xF$ is equivalent to $FR_xF = R_x$. But $FR_xF = R_x^{-1}$.
So $R_x = R_x^{-1}$. The only rotation that satisfies this equation is $R_{180}$. So the only rotation that $C(F)$ can contain is $R_{180}$.

We may conclude that for even $n$, $C(F) = \lbrace R_0, R_{180}, F, F_0 \rbrace$ where $FF_0 = R_{180}$ and for odd $n$, $C(F) = \lbrace R_0, F \rbrace$.

Next we find $C(R)$. Observe that $RR_x = R_xR$ for all rotations $R_x$ in $D_n$. Let $F_1$ be a reflection in $D_n$. Then $RF_1 = F_1R$ is equivalent to $F_1RF_1 = R$ which implies $R = R^{-1}$. This equation is true only when $R = R_{180}$. Thus, $C(R_{180}) = D_n$. For $R \neq R_{180}$, we have $C(R) = \lbrace R_0, R_{\frac{360}{n}}, R_{2\cdot\frac{360}{n}}, \dots, R_{(n-1)\cdot\frac{360}{n}} \rbrace$.

Can someone please check this?

 

[fresh_42]

I think this is way too complicated and I have difficulties to follow you. $F_0$ is disturbing, to say the least. The dihedral groups have a nice normal form for its elements: we can always write any element as $R^kF^\varepsilon$ with $k\in \mathbb{Z}$ and $\varepsilon \in \{\,0,1\,\}\,.$

So the question is: When is $R^kF^\varepsilon \in C(F)\, , \,C(R)$ resp.? With the other relations $F^2=R^n=1$ this can be solved more easily.

 

[fishturtle1]

I think this is way too complicated and I have difficulties to follow you. $F_0$ is disturbing, to say the least. The dihedral groups have a nice normal form for its elements: we can always write any element as $R^kF^\varepsilon$ with $k\in \mathbb{Z}$ and $\varepsilon \in \{\,0,1\,\}\,.$

So the question is: When is $R^kF^\varepsilon \in C(F)\, , \,C(R)$ resp.? With the other relations $F^2=R^n=1$ this can be solved more easily.

Thank you for the feedback. I tried to write a simpler solution with what you mention, but I am not sure I got there...

Let $D_n$ be a dihedral group and fix a reflection $F$ and rotation $R$ in $D_n$. Observe that every element in $D_n$ can be written as $R^kF^\varepsilon$ for some integer $k$ and $\varepsilon \in \lbrace 0, 1 \rbrace$.

First we find $C(F)$. Suppose $R^kF^\varepsilon$ that commutes with $F$ and consider 2 cases.

case 1: $\varepsilon = 1$. Then $(R^kF)F = F(R^kF)$. This simplifies to $R^k = FR^kF$. But from relevant equations, this implies that $R^k = R^{-k}$. The only rotation that satisfies this is $R^{\frac{n-1}{2}}$(the rotation of 180 degrees). This implies that the only reflections that can be in $C(F)$ is $R^{\frac{n-1}{2}}F$.

case2: $\varepsilon = 0$. Then $R^kF = FR^k$. This implies $FR^kF = R^k$. By relevant equations, this implies $R^k = R^{-k}$. The only rotation that satisfies this is $R^{\frac{n-1}{2}}$. So the only rotation that can be in $C(F)$ is $R^{\frac{n-1}{2}}$.

In conclusion, for even $n$, $C(F) = \lbrace R^0, R^{\frac{n-1}{2}}, R^{\frac{n-1}{2}}F, F\rbrace$ and for odd $n$, $C(F) = \lbrace R^0, F \rbrace$.

Next we find $C(R)$. Suppose $R^kF^\varepsilon$ commutes with $R$ and consider 2 cases:

case1: $\varepsilon = 1$. Then $R(R^kF) = (R^kF)R$. This simplifies to $R^{k+1}F = R^kFR$. Multiplying on the left of both sides by $R^{-k}$, we get $RF = FR$. Multiplying on the left of both sides by $F$, we get $FRF = R$. By relevant equations this implies $R = R^{-1}$. So we must have $R = R^{\frac{n-1}{2}}$. This implies? the only reflection that commutes with $R$ is $R^{\frac{n-1}{2}}F$.

case2: $\varepsilon = 0$. Then $RR^k = R^kR$ which is satisfied by all rotations.

So $C(R) = \lbrace R^0, R^1, R^2, \dots, R^{n-1}, R^{\frac{n-1}{2}}F \rbrace$.

But I'm pretty sure this is wrong, since in another thread it says that the center of $D_n$ is $Z(D_n) = \lbrace R^0, R^{\frac{n-1}{2}} \rbrace$ for even $n$. But here we get $Z(D_n) = \lbrace R^0, R^{\frac{n-1}{2}}, R^{\frac{n-1}{2}}F \rbrace$ for even $n$.

 

[fishturtle1]

let me try this again...

 

[fresh_42]

Thank you for the feedback. I tried to write a simpler solution with what you mention, but I am not sure I got there...

Let $D_n$ be a dihedral group and fix a reflection $F$ and rotation $R$ in $D_n$. Observe that every element in $D_n$ can be written as $R^kF^\varepsilon$ for some integer $k$ and $\varepsilon \in \lbrace 0, 1 \rbrace$.

First we find $C(F)$. Suppose $R^kF^\varepsilon$ that commutes with $F$ and consider 2 cases.

case 1: $\varepsilon = 1$. Then $(R^kF)F = F(R^kF)$. This simplifies to $R^k = FR^kF$. But from relevant equations, this implies that $R^k = R^{-k}$. The only rotation that satisfies this is $R^{\frac{n-1}{2}}$(the rotation of 180 degrees). This implies that the only reflections that can be in $C(F)$ is $R^{\frac{n-1}{2}}F$.

case2: $\varepsilon = 0$. Then $R^kF = FR^k$. This implies $FR^kF = R^k$. By relevant equations, this implies $R^k = R^{-k}$. The only rotation that satisfies this is $R^{\frac{n-1}{2}}$. So the only rotation that can be in $C(F)$ is $R^{\frac{n-1}{2}}$.

In conclusion, for even $n$, $C(F) = \lbrace R^0, R^{\frac{n-1}{2}}, R^{\frac{n-1}{2}}F, F\rbrace$ and for odd $n$, $C(F) = \lbrace R^0, F \rbrace$.

Where do you get $n-1$ from?

Yes, we have $R^k=R^{-k}$ in both cases. This means $R^{2k}=1$, i.e. $n\,|\,2k$, since $R^n=1$ is the smallest natural number, the order of $R$. If $n$ is odd, then $n\,|\,k$ and $R^k=1$. So $C_{D_n}(F)=\{\,F,1\,\}\cong \mathbb{Z}_2$ for odd $n$. For even $n=2m$ we get $m\,|\,k$, say $m\cdot c= k$. Then $R^k=R^{mc}=R^{\frac{n}{2}c}$ and we must test, whether all these elements ($c \in \mathbb{Z}$ arbitrary) actually commute with$F$, because we only have a necessary, no sufficient condition. (The test for odd $n$ is trivial.)

Next we find $C(R)$. Suppose $R^kF^\varepsilon$ commutes with $R$ and consider 2 cases:

case1: $\varepsilon = 1$. Then $R(R^kF) = (R^kF)R$. This simplifies to $R^{k+1}F = R^kFR$. Multiplying on the left of both sides by $R^{-k}$, we get $RF = FR$. Multiplying on the left of both sides by $F$, we get $FRF = R$. By relevant equations this implies $R = R^{-1}$. So we must have $R = R^{\frac{n-1}{2}}$. This implies? the only reflection that commutes with $R$ is $R^{\frac{n-1}{2}}F$.

case2: $\varepsilon = 0$. Then $RR^k = R^kR$ which is satisfied by all rotations.

So $C(R) = \lbrace R^0, R^1, R^2, \dots, R^{n-1}, R^{\frac{n-1}{2}}F \rbrace$.

But I'm pretty sure this is wrong, since in another thread it says that the center of $D_n$ is $Z(D_n) = \lbrace R^0, R^{\frac{n-1}{2}} \rbrace$ for even $n$. But here we get $Z(D_n) = \lbrace R^0, R^{\frac{n-1}{2}}, R^{\frac{n-1}{2}}F \rbrace$ for even $n$.

Same problem. It is correct that we have $R=R^{-1}$. This means $R^2=1$ and $n\,|\,2$, which is only possible for $n=2$. So $C_{D_2}(R)=\{\,1,R,RF,F\,\}=D_2$ which makes sense as $D_2$ is Abelian. The element $F$ in the centralizer corresponds to the case $k=0\,.$ For $n>2$ we get $C_{D_n}(R)=\{\,1,R,\ldots,R^{n-1}\,\}=\langle R \rangle \cong \mathbb{Z}_n$ because $R^2=1$ is a contradiction, i.e. $\varepsilon=0\,.$

 

[fishturtle1]

Where do you get $n-1$ from?

Yes, we have $R^k=R^{-k}$ in both cases. This means $R^{2k}=1$, i.e. $n\,|\,2k$, since $R^n=1$ is the smallest natural number, the order of $R$. If $n$ is odd, then $n\,|\,k$ and $R^k=1$. So $C_{D_n}(F)=\{\,F,1\,\}\cong \mathbb{Z}_2$ for odd $n$. For even $n=2m$ we get $m\,|\,k$, say $m\cdot c= k$. Then $R^k=R^{mc}=R^{\frac{n}{2}c}$ and we must test, whether all these elements ($c \in \mathbb{Z}$ arbitrary) actually commute with$F$, because we only have a necessary, no sufficient condition. (The test for odd $n$ is trivial.)

The $n-1$ is my mistake. Instead of $R^{\frac{n-1}{2}}$ I meant $R_{180}$.

Ok so just focussing on $C_{D_n}(F)$,

Let $R^kF^\varepsilon \in C(F)$. This means $(R^kF^\varepsilon)F = F(R^kF\varepsilon)$. In either case of $\varepsilon = 0$ or $\varepsilon = 1$, we get $R^k = R^{-k}$. Thus $R^{2k} = 1$. Since $\vert R \vert = n$, we have $n \vert 2k$.

If $n$ is odd, then $n \vert k$. Thus $R^k = 1$. So the possible elements of $C(F)$ are $F$ and $1$. Clearly, $F$ commutes with $F$ and $F1 = 1F = F$. So, for odd $n$, we have $C_{D_n}(F) = \lbrace 1 , F \rbrace$.

Suppose $n$ is even. Then $n = 2m$ for some integer $m$. We have $n \vert 2k$ which implies $m\cdot c = k$ for some integer $c$. So $R^k = R^{m\cdot c} = R^{\frac n2c}$. So we need to find which rotations of the form $R^{\frac n2c}$ satisfies $(R^{\frac n2c}F^\varepsilon)F = F(R^{\frac n2c}F^\varepsilon)$. But how do I check this? it doesn't seem like $R^{\frac n2c}$ is any specific rotation?

edit: So if $R^kF^\varepsilon \in C_{D_n}(F)$ then $R^kF^\varepsilon \in \lbrace R^{\frac2n c}F, R^{\frac2n c} \rbrace$. So I need to check which of these two elements commute with $F$.

 

[fresh_42]

The $n-1$ is my mistake. Instead of $R^{\frac{n-1}{2}}$ I meant $R_{180}$.

Ok so just focussing on $C_{D_n}(F)$,

Let $R^kF^\varepsilon \in C(F)$. This means $(R^kF^\varepsilon)F = F(R^kF\varepsilon)$. In either case of $\varepsilon = 0$ or $\varepsilon = 1$, we get $R^k = R^{-k}$. Thus $R^{2k} = 1$. Since $\vert R \vert = n$, we have $n \vert 2k$.

If $n$ is odd, then $n \vert k$. Thus $R^k = 1$. So the possible elements of $C(F)$ are $F$ and $1$. Clearly, $F$ commutes with $F$ and $F1 = 1F = F$. So, for odd $n$, we have $C_{D_n}(F) = \lbrace 1 , F \rbrace$.

Suppose $n$ is even. Then $n = 2m$ for some integer $m$. We have $n \vert 2k$ which implies $m\cdot c = k$ for some integer $c$. So $R^k = R^{m\cdot c} = R^{\frac n2c}$. So we need to find which rotations of the form $R^{\frac n2c}$ satisfies $(R^{\frac n2c}F^\varepsilon)F = F(R^{\frac n2c}F^\varepsilon)$. But how do I check this? it doesn't seem like $R^{\frac n2c}$ is any specific rotation?

edit: So if $R^kF^\varepsilon \in C_{D_n}(F)$ then $R^kF^\varepsilon \in \lbrace R^{\frac2n c}F, R^{\frac2n c} \rbrace$. So I need to check which of these two elements commute with $F$.

Don't forget that $F\in C_{D_n}(F)$.

If $R^k\in C_{D_n}(F)$ then we have $R^kF \in C_{D_n}(F)$ so we can concentrate on pure rotations $R^k$. We have shown that $x=R^k\in C_{D_n}(F)$ implies $k \in \mathbb{Z}\cdot \frac{n}{2}$.
$$
R^{\frac{n}{2}c}\cdot F = F \cdot R^{-\frac{n}{2}c} = F \cdot 1 \cdot R^{-\frac{n}{2}c} = F \cdot R^{nc}\cdot R^{-\frac{n}{2}c} = \ldots
$$
... and we only have to gather what is there. By the way, how many elements do we get out of $R^{\frac{n}{2}c}$ with arbitrary integers $c$?

 

[fishturtle1]

Don't forget that $F\in C_{D_n}(F)$.

If $R^k\in C_{D_n}(F)$ then we have $R^kF \in C_{D_n}(F)$ so we can concentrate on pure rotations $R^k$. We have shown that $x=R^k\in C_{D_n}(F)$ implies $k \in \mathbb{Z}\cdot \frac{n}{2}$.
$$
R^{\frac{n}{2}c}\cdot F = F \cdot R^{-\frac{n}{2}c} = F \cdot 1 \cdot R^{-\frac{n}{2}c} = F \cdot R^{nc}\cdot R^{-\frac{n}{2}c} = \ldots
$$
... and we only have to gather what is there. By the way, how many elements do we get out of $R^{\frac{n}{2}c}$ with arbitrary integers $c$?

So continuing from above, we have
$$
R^{\frac{n}{2}c}\cdot F = F \cdot R^{-\frac{n}{2}c} = F \cdot 1 \cdot R^{-\frac{n}{2}c} = F \cdot R^{nc}\cdot R^{-\frac{n}{2}c} = FR^{nc - \frac{n}{2}c} = FR^{\frac n2c}
$$

So $R^{\frac n2c} \in C_{D_n}(F)$. Since $F \in C_{D_n}(F)$, we have $R^{\frac n2c}F \in C_{D_n}(F)$.

Moreover, for even $c$, $R^{\frac n2c} = 1$ and for odd $c$, $R^{\frac n2c} = R^{\frac n2}$. Also, $(R^{\frac n2})^2 = 1$ which implies $FR^{\frac n2}F = R^{\frac n2}$ which implies $FR^{\frac n2} = R^{\frac n2}F$.

Thus, $C_{D_n}(F) = \lbrace 1, R^{\frac n2}, F, R^{\frac n2}F \rbrace$.

 

[fishturtle1]

Same problem. It is correct that we have $R=R^{-1}$. This means $R^2=1$ and $n\,|\,2$, which is only possible for $n=2$. So $C_{D_2}(R)=\{\,1,R,RF,F\,\}=D_2$ which makes sense as $D_2$ is Abelian. The element $F$ in the centralizer corresponds to the case $k=0\,.$ For $n>2$ we get $C_{D_n}(R)=\{\,1,R,\ldots,R^{n-1}\,\}=\langle R \rangle \cong \mathbb{Z}_n$ because $R^2=1$ is a contradiction, i.e. $\varepsilon=0\,.$

We said earlier that any element in $D_n$ can be written as $R^kF$. But this would require $<R>$ is the set of all rotations of $D_n$, right? So I think when finding $C(R)$, we should say $R := R_{\frac{360}{n}}$ and then solve for $C(R^l)$ for arbitrary $l$. Is this ok?

So to find $C(R^l)$... Let $R^kF^\varepsilon \in C(R^l)$. Clearly $R^kF^\varepsilon \in C(R^l)$ if $\varepsilon = 0$. Suppose that $\varepsilon = 1$. Then $(R^kF)R^l = R^l(R^kF)$ implies $FR^l = R^lF$ which implies $R^l = R^{-l}$ i.e. $R^{2l} = 1$. This means $\vert R^l \vert \le 2$. This implies $R^l = 1$ or $R^l = R^{\frac n2}$. Therefore, we need only test if $R^kF \in C(1)$ and if $R^kF \in C(R^{\frac n2})$.

Clearly $R^kF \in C(1) = D_n$.

We know $FR^{\frac n2}F = R^{-\frac n2}$. Multiplying both sides by $R^n$, we have $FR^{\frac n2}F = R^{-\frac n2}R^n = R^{\frac n2}$. Multiplying on the right of both sides by $F$, we get $FR^{\frac n2}F^2 = R^{\frac n2}F$ which simplifies to $FR^{\frac n2} = R^{\frac n2}F$. Multiplying on the left of both sides by $R^k$, we get $R^kFR^{\frac n2} = R^kR^{\frac n2}F$ which can be rewritten as $(R^kF)R^{\frac n2} = R^{\frac n2}(R^kF)$. We may conclude $R^kF \in C(R^{\frac n2})$. Since $k$ was arbitrary, we have $C(R^l) = D_n$.

Lastly, for $R^l \neq 1, R^{\frac n2}$, we have $C(R^l) = <R>$. []

 

[fresh_42]

I'm a bit confused because you change horses in the middle of the race.
You started with an arbitrary element $X=R^kF\in C_{D_n}(R^l)$, which means that $l$ is a given integer and we want to find out, what this means for $k$.
But then you say $l=\frac{n}{2}$.
Decide which horse you want to ride: given $l$, or given the fact that there is an element $X$ at all in some $C_{D_n}(R^l)$.

I assume that we still want to calculate $C_{D_n}(R^l)$. We already know that $\langle R \rangle \subseteq C_{D_n}(R^l)$. From $R^kF \in C_{D_n}(R^l)$ we get $R^{2l}=1$ and $|R^l|\leq 2$, correct. Better would be to say that $n\,|\,2l$. If $n$ is odd, then $n\,|\,l$ or $X\notin C_{D_n}(R^l)$; if $n$ is even, then $\frac{n}{2}\,|\,l$ or $X\notin C_{D_n}(R^l)$.
$$
C_{D_n}(R^l) = \begin{cases} \langle R \,\rangle &\quad \text{ if } n \text{ is odd and } n \nmid l \\ D_n &\quad \text{ if }n \text{ is odd and } n \,|\, l \text{ since } R^l=1\\ \langle R \,\rangle &\quad \text{ if } n \text{ is even and }\frac{n}{2} \nmid l \\D_n& \quad \text{ if } n \text{ is even and } \frac{n}{2} \,|\, l \text{ with } R^l\in \{\,1\, , \,R^{n/2}\,\}\end{cases}
$$
You have all components, but as mentioned your way is a bit confusing. We have a given $n$ and a given $l$, and a variable $X$ which might or might not exist in $C_{D_n}(R^l)$.

 

[fishturtle1]

I'm a bit confused because you change horses in the middle of the race.
You started with an arbitrary element $X=R^kF\in C_{D_n}(R^l)$, which means that $l$ is a given integer and we want to find out, what this means for $k$.
But then you say $l=\frac{n}{2}$.
Decide which horse you want to ride: given $l$, or given the fact that there is an element $X$ at all in some $C_{D_n}(R^l)$.

I assume that we still want to calculate $C_{D_n}(R^l)$. We already know that $\langle R \rangle \subseteq C_{D_n}(R^l)$. From $R^kF \in C_{D_n}(R^l)$ we get $R^{2l}=1$ and $|R^l|\leq 2$, correct. Better would be to say that $n\,|\,2l$. If $n$ is odd, then $n\,|\,l$ or $X\notin C_{D_n}(R^l)$; if $n$ is even, then $\frac{n}{2}\,|\,l$ or $X\notin C_{D_n}(R^l)$.
$$
C_{D_n}(R^l) = \begin{cases} \langle R \,\rangle &\quad \text{ if } n \text{ is odd and } n \nmid l \\ D_n &\quad \text{ if }n \text{ is odd and } n \,|\, l \text{ since } R^l=1\\ \langle R \,\rangle &\quad \text{ if } n \text{ is even and }\frac{n}{2} \nmid l \\D_n& \quad \text{ if } n \text{ is even and } \frac{n}{2} \,|\, l \text{ with } R^l\in \{\,1\, , \,R^{n/2}\,\}\end{cases}
$$
You have all components, but as mentioned your way is a bit confusing. We have a given $n$ and a given $l$, and a variable $X$ which might or might not exist in $C_{D_n}(R^l)$.

That's my mistake, $R^l$ is supposed to be fixed, especially because the problem statement says so. But I did not treat it as fixed, as you pointed out. Let me try again using what you've said. Thank you for your time on this.

 

[fishturtle1]

Let $R := R_{\frac{360}{n}}$ in a the dihedral group $D_n$. Fix some integer $l$. We will calculate $C_{D_n}(R^l)$. Clearly, $<R> \subseteq C_{D_n}(R^l)$.

Suppose $X = R^kF \in C_{D_n}(R^l)$ for some integer $k$. Then $XR^l = R^lX$ implies $R^{2l} = 1$. Thus, $n \vert 2l$.

Suppose $n$ is odd. If $n \nmid l$ then $n \nmid 2l$. This implies $XR^l \neq R^lX$. Thus, if $n \nmid l$ and $n$ is odd, then $C_{D_n}(R^l) = <R>$. If $n$ is odd and $n \vert l$, then $R^l = 1$ which implies $C_{D_n}(R^l) = D_n$.

Suppose $n$ is even. Suppose $n \vert 2l$. This is equivalent to $\frac n2 \vert l$ which is equivalent to $l = \frac n2 \cdot c$ for some integer $c$. We have $XR^lX = R^{-l}$ which is equivalent to $XR^{\frac n2c}X = R^{-\frac n2c}$. This implies $XR^{\frac n2c} = R^{-\frac n2c}X = 1\cdot R^{-\frac n2c}X = R^{nc}R^{-\frac n2c}X = R^{\frac n2 c}X$. Thus $XR^l = R^lX$. We can conclude if $n$ is even and $n \vert 2l$ then $C_{D_n}(R^l) = D_n$.

Lastly, suppose $n$ is even and $n \nmid 2l$. Then $R^{2l} \neq 1$. This implies $XR^l \neq R^lX$. Thus, if $n$ is even and $n \nmid 2l$ then $C_{D_n}(R^l) = <R>$