# Number of elements of order 2 in $D_n/Z(D_n)$

This exercise in supplimentry exercises for chapter 9 to 11 of contemporary Abstract algebra book by Gallian

Let $$n=2m$$, where $$m$$ is odd. How many elements of order $$2$$ does the group $$D_n/Z(D_n)$$ have?

Answer of this exercise given ( in "selected answer" section) is: $$m$$ elements

Author used the notation $$D_n$$ (everywhere in book) to represent Dihedral group of order $$2n$$. But according to me, if we use this notation(by author) then given answer is wrong!

(According to me, if we use the notation $$D_{2m}$$ for Dihedral group of order $$2m$$ then answer is true! that is we consider the group $$D_{2m}$$ Dihedral group of order $$2m$$; where $$m$$ is odd then number of elements of order $$2$$ in $$D_{2m}/Z(D_{2m})$$ is $$m$$ is true! )

• I hate that $|D_n|=n$ is even a possibility people have to consider. It leads to so much confusion, like here (seems the author confused themselves). Why can't we all just agree that $D_n$ (at least for $n\geq3$) should act on $n$ elements the way $S_n$ and $A_n$ do, and thus $|D_n|=2n$? Commented May 31, 2021 at 6:08
• @Arthur, completely agrees with you. Commented May 31, 2021 at 9:54
• @Akash Patalwanshi please check my edited answer. Commented Jun 6, 2021 at 7:32

No , you are wrong.

In both case , answer will be same , i.e, there are $$m$$ elements of order $$2$$ in $$D_{n}/Z(D_n)$$

Edit : If $$|D_n|=2n$$ and $$n=2m$$, where $$m$$ is odd.

Then , it is true that there are $$m$$ elements of order $$2$$ in $$D_n/Z(D_n)$$.

Because, $$D_{n}/Z(D_n) \cong D_m$$ $$\text{(here, |D_m|=2m)}$$

(How? , Clearly $$Z(D_n)=\{1,r^m\}$$ where , $$D_n=\{1,r,r^2,\cdot\cdot\cdot,r^{2m-1},s,sr,sr^{2},\cdot\cdot\cdot,sr^{2m-1}\}$$

Then , $$1Z(D_n)=\{1,r^m\}=r^{m} Z(D_n)$$ $$rZ(D_n)=\{r,r^{m+1}\}=r^{m+1} Z(D_n)$$

$$\cdot\cdot\cdot\cdot$$

$$\cdot\cdot\cdot\cdot$$

$$r^{m-1}Z(D_n)=\{r^{m-1},r^{2m-1}\}=r^{2m-1} Z(D_n)$$

Similarly, $$sZ(D_n)=\{s,sr^{m}\}=sr^{m+1} Z(D_n)$$

$$\cdot\cdot\cdot\cdot$$

$$\cdot\cdot\cdot\cdot$$

$$sr^{m-1}Z(D_n)=\{sr^{m-1},sr^{2m-1}\}=sr^{2m-1} Z(D_n)$$

So, $$D_{n}/Z(D_n)= \{Z(D_n),rZ(D_n),\cdot\cdot\cdot,r^{m-1}Z(D_n),sZ(D_n),srZ(D_n),\cdot\cdot\cdot,sr^{m-1}Z(D_n)\} \cong D_m$$)

If $$|D_n|=n$$ and $$n=2m$$, where $$m$$ is odd.

Then $$Z(D_n)$$ is always trivial, so, $$D_n/Z(D_n) \cong D_n$$ , and there are $$m$$ reflection symmetries in $$m-$$ gon.

• Thank you so much sir. Commented May 31, 2021 at 10:16
• I don't understand how $D_4$ is a counterexample in the first part of your answer because $m = 2$ is not odd. Commented Jun 4, 2021 at 22:34
• @Zoe H yeah you're right, I have edited. Commented Jun 6, 2021 at 7:31

n=2m and m is odd, here Z(D_n) consists of two elements that is {R_0,R_180} and since m is odd then no other elements such R_90,R_270 can exist in Z(D_n) And since the reflection are distinct then the no of elements of order 2 is just n/3=m.