Question

Find Z(D2n), where D2n is the dihedral group with 2n elements,

i) when n is an odd integer;

ii) when n is an even integer.

Answer 1

i)

$Z(D_{2n})=1$

ii)

$Z(D_{2n})=\{1,r^k\}$

r and s generate D_2n with the group presentation $\{r,s|r^n=s^2=1,rs=sr^{-1\}}$

Proof:

i)

For any $x\isin Z(D_{2n})$, x must commute with both s and r, since both are in D_2n. If x

commutes with both s and r, then x commutes with any element of D_2n since r and s generate D_2n.

Then for any $x\isin Z(D_{2n})$, we have xr=rx, where x is of the form x=s^jr^w. Since r and s have finite order, the exponents are modulo n and modulo 2 for rand s respectively.

Then we have s^jr^wr=rs^jr^w$\implies$ s^jr^w+1=s^jr^w-1

In the case where the power of r on the RHS is w-1, it is clear we cannot have equality unless

|r|=1, which it is not. If j is even then we have only the w+1 case. Hence x may be of the form

1r^w=r^w.

Also we have that x must commute with s, hence sx=xs.

Then sr^w=r^ws$\implies$ sr^w=sr^-w

Applying s^-1 to both sides we arrive at the task of finding when r^w=r^-w i.e when w=-w. But since r has finite order we have: 

$\overline{w}=(n-1)w$

where $\overline{w}$ is the residue class of w modulo n. Hence for some $a\isin Z_+$

w+an=wn-w

2w=wn-an

2w=n(w-a)

Since 0<w<n we have that the LHS is greater than 0. Also since $n\ge w$ then $(w-a)\le 2$ ,

since otherwise the RHS would be greater than the LHS. So since n is odd we must have that

(w-a) is even and hence (w-a)=2, implying that 2w=2n and hence n=w.

Hence x=r^w=rⁿ=1. So Z(D_{2n})=1

ii)

From the argument above any $x\isin Z(D_{2n})$  must be of the form s^jr^w for j even and $j\isin Z$ .

Then x is of the form r^w for $0<w\le n$ . Since n is even we have the equation below is satisfied when 2w=n or w=n, since $(w-a)\le 2$ but also $(w-a)\ge 0$ for $(w-a)\isin Z_+$

2w=n(w-a)

Hence $r^w\isin Z(D_{2n})$ when 2w=n, as desired.