Center of $D_6$ is $\mathbb{Z}_2$

Question

The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.

I have that $$D_6=\left< a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right>$$ $$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$ My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that $$a^3b=ba^3$$ and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?

For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.

Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible.

Answer 1

I wrote up a general classification for the centers of $D_n$, (the dihedral group of order $2n$, not $n$) just the other week. Perhaps it will be useful to read:

If $n=1,2$, then $D_n$ is of order $2$ or $4$, hence abelian, and $Z(D_n)=D_n$. Suppose $n\geq 3$. We have the presentation $$ D_n=\langle x,y:x^2=y^n=1,\; xyx=y^{-1}\rangle. $$ Then $yx=xy^{-1}$ implies the reduction $y^kx=xy^{-k}$. An element is in the center iff it commutes with $x$ and $y$, since $x$ and $y$ generate $D_n$. Let $z=x^iy^j$ be in the center. From $zy=yz$ we see $$ x^iy^{j+1}=yx^iy^j\implies x^iy=yx^i. $$ But $i\neq 1$, else we have $xy=yx=xy^{-1}$, so $y^2=1$, a contradiction since $n\geq 3$. So $i=0$, and $z=y^j$. Then from the equation $zx=xz$, we have $$ y^jx=xy^j=xy^{-j} $$ which implies $y^{2j}=1$. Thus $j=0$ or $j=n/2$. If $n$ is odd, we must necessarily have $j=0$, and $z=1$. If $n$ is even, either possibility works. But $y^{n/2}$ is indeed in the center as it clearly commutes with $y$, as well as with $x$ since $y^{n/2}x=xy^{-n/2}=x(y^{n/2})^{-1}=xy^{n/2}$. Summarizing, we have, for $n\geq 3$, $$ Z(D_n)=\begin{cases} \{1,y^{n/2}\} & \text{if }n\equiv 0\pmod{2},\\ \{1\} & \text{if }n\equiv 1\pmod{2}. \end{cases} $$

Answer 2

General method to find the centre of $D_{2n}$

if $n=1$ or $n=2$ ,then $D_{2n}$ is abelian and hence $Z(D_{2n})= D_{2n}.$

Now suppose $ n \ge 3$ .By definition we have

$D_{2n} = \{ a^ib^j :i=0,1 , j=0,1,....., n-1\}$ where $a$ is an element of order $2$, $b$ is an element of order $n$ and $a, b$ are given by relation $ ba=ab^{-1}$

Then its implies $bba=b^2a = bab^{-1}=ab^{-1}b^{-1}=ab^{-2}.$

In general form we can write

$b^ra=ab^{-r}\tag1$

For all integer $r \ge 0$.Now , since $a$ and $b$ together generate $D_{2n}$, an element of $D_{2n}$ is in center if and only if its commutes with both $a$ and $b$.

Therefore $ x= a^ib^j \in Z(D_{2n})$ if and only if $xa=ax$ and $xb=bx$.

$x b= a^ib^j b=a^ib^{j+1}= bx= ba^ib^j$ where $x= a^ib^j$

Therefore $a^ib=ba^i \tag 2$

we can see clearly $i=0$ in $(2)$ because $a^0b=ba^0$ implies $ b=b$

we can see clearly $i \neq 1$ because if $i=1$ then $ab=ba$ but from $(1)$ , for $n \ge 3$ we have $ab=ba^{-1} \neq ba$ so we are getting contradiction if $i=1$

we have have $x=a^ib^j$ . If $i=0$ then $x= a^0b^j=b^j$

Now the condition $xa=ax$ t becomes $b^ja=ab^j$

from $(1)$ relation we have $ab^{-j}=ab^j$

There $ b^{-j}=b^j$ implies $b^{2j}=1 \tag 3$

Since order$(b)=n$ (given), we get from $(3)$ that $$n |2j$$ Therefore if $n$ divide $2j$ then $j=0$ or $2j=n$

Now if $j=0$ then $x=b^j=b^0=1$

If $2j=n$ , then $n$ is even , that is $j= \frac{n}{2}$

so $x=b^j=b^{n/2}$

so we have proved that $$Z(D_{2n})= \begin{cases} D_{2n} \ \text{if n=1 ,2} \\ \{1\}\ \text{if n > 2 is odd} \\ \{1,b^{n/2}\} \ \text{if n >2 is even } \end{cases}$$