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! )
Is, am I correct? Please help.
 A: 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.
A: 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.
