Answer to Question #203308 in Abstract Algebra for Raghav

Question #203308

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


1
Expert's answer
2021-06-14T14:20:37-0400

i)

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


ii)

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

r and s generate D2n 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 D2n.  If x

commutes with both s and r, then x commutes with any element of D2n since r and s generate D2n.

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

Then we have sjrwr=rsjrw"\\implies" sjrw+1=sjrw-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

1rw=rw.

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

Then srw=rws"\\implies" srw=sr-w

Applying s-1 to both sides we arrive at the task of finding when rw=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=rw=rn=1. So Z(D_{2n})=1


ii)

From the argument above any "x\\isin Z(D_{2n})"  must be of the form sjrw for j even and "j\\isin Z" .

Then x is of the form rw 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.


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