Center of dihedral groups

Posted: February 2, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

Problem. Let n \geq 1 and let D_{2n} be the dihedral group of order 2n. Find Z(D_{2n}), the center of D_{2n}.

Solution. If n=1 or n=2, then D_{2n} is abelian and hence Z(D_{2n})=D_{2n}. Now, suppose n \geq 3. By definition, we have D_{2n}=\{a^ib^j: \ i=0,1, \ j=0,1, \cdots , n-1 \}, where a is an element of order 2, b is an element of order n and a,b are related by the relation ba=ab^{-1}. It then follows that b^2 a= bab^{-1}=ab^{-2} and in general

b^ra=ab^{-r}, \ \ \ \ \ \ \ \ \ \ (1)

for all integers r \geq 0. Now, since a and b together generate D_{2n}, an element of D_{2n} is in the center if and only if it commutes with both a and b. So x=a^ib^j \in Z(D_{2n}) if and only if xa=ax and xb=bx. The condition xb=bx gives us a^ib^{j+1}=ba^ib^j and thus

a^ib=ba^i. \ \ \ \ \ \ \ \ \ \ (2)

Clearly we can have i=0 in (2) but can we have i=1 ? The answer is no because if i=1, then (2) gives ab=ba and we know from (1) that ba=ab^{-1}. But then ab=ab^{-1} and so b^2=1, which is not possible because n \geq 3. So i=0 and thus x=b^j. The condition xa=ax then becomes b^ja=ab^j and hence, by (1), ab^{-j}=ab^j. Therefore

b^{2j}=1. \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

Hence, since o(b)=n, we get from (3) that n \mid 2j. Thus either j=0 or 2j=n because 0 \leq j \leq n-1. If j=0, then x=b^j=1. If 2j=n, then n is even and x=b^j = b^{n/2}. So we have proved

Z(D_{2n}) = \begin{cases} D_{2n} & \text{if} \ n=1,2 \\ \{1\} & \text{if} \ n > 2 \ \text{is odd} \\ \{1,b^{n/2} \} & \text{if} \ n > 2 \ \text{is even}. \ \Box \end{cases}

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Comments
  1. That guy says:

    Thanks mate, good stuff, easy to follow

  2. emi says:

    really great

  3. Meena Pargaei says:

    this is very good proof…

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