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

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

I don't how to begin this proof. All I have so far is that Dn/Z(Dn) should have one element of order 2.

• The first thing you have to do is work out what $Z(D_n)$ is. Can you do that? Also, $D_n/Z(D_n)$ is not a subgroup, it's a quotient group. – Gerry Myerson Dec 8 '13 at 3:42
• would it be Z? just the integers – Ash Dec 8 '13 at 3:54
• I think you misunderstand the notation. The integers are not involved. If $G$ is a group, $Z(G)$ is notation for the center of the group. So you need to know what $D_n$ means, and then you need to figure out what its center is. – Gerry Myerson Dec 8 '13 at 3:56
• I know Dn is the Dihedral group and for D4 the center is {Ro,R180} but im not sure what the center for Dn is. would it just be the Dihedral group without R0? – Ash Dec 8 '13 at 4:15
• Huh? if by $R_0$ you mean the identity element of $D_n$, then that's always in the center. Do you know what "center" means in the context of group theory? – Gerry Myerson Dec 8 '13 at 4:31

Hint: notice that n = 2m where m is odd that implies $D_n$ has a sylow 2- subgroup.
Now see that $D_n$ is a finite group and $Z(D_n)$ the center of $D_n$ is a subgroup of $D_n$ then what can you say about the order and index of these groups and subgroups, also use Lagrange's theorem.