# Center of R, F in D_n

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##.

fresh_42
Mentor
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##.

let me try this again...

fresh_42
Mentor
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
Mentor
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##.

• fresh_42
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
Mentor
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.

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>##