## Center of dihedral groups

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

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

1. Ben irwin says:

Hi, do you not mean to conclude that the centre for an even n would be {1, a^(n/2)}?

In D12 for example, we have a^3b =ba^3 and not b^3a=ab^3 as b^3a=a^5b.

• Yaghoub Sharifi says:

No, the center of $D_{12}$ is $\{1,b^3\}.$ Notice that, with my notation, in $D_{12}, \ a$ has order 2 and $b$ has order 6 (you’re probably assuming $a$ has order 6 and $b$ has order 2). So we can’t have $a^3b=ba^3$ because then $D_{12}$ would be abelian (because $a^3=a$).

In the very end, for even n, centre should be {1,r^n} instead of {1,r^(n/2)} which is correct if you are using D_n notation but rather you are using D_2n notation.

• Yaghoub Sharifi says:

No, $b^{n/2}$ is correct; read the solution again.

• Jesus says:

Bhaskar is correct. D_8 implies n=4 and n/2=2, b^2 it’s not in the center but b^4 is.
b^2(b^3a)=b^5a
(b^3a)b^2=b^3(b^6a)=ba
Therefore, b^2 is not in the center.

• Yaghoub Sharifi says:

No, the center of $D_8,$ as I have proved, is $\{1,b^2\}.$
Remember that in $D_8$ the order of $b$ is 4. So $b^4=1,$ which is obviously in the center.
Also $b^5a=ba$ because $b^5=b.$

3. Subhankar Saha says:

This is a very useful proof.

4. That guy says:

Thanks mate, good stuff, easy to follow

5. emi says:

really great

6. Meena Pargaei says:

this is very good proof…