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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 $\mathbb{Z}_2$.

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

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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}

$$

It does not give proofs, but given the tone of the original question a good graphical tool for the OP would be Group Explorer. It does a lot of the donkey work, and can show you various visualisations, including the multiplication tables in helpful ways.

http://groupexplorer.sourceforge.net/

For $D_6$, that $Z_2$ is the center is (to me a least) kinda obvious by just looking at the pictures of the multication table sorted by it’s various subgroups. eg below is $D_6$ with the required subgroup shown in top left – note the first 2 rows and 2 columns match exactly, whereas other rows/columns don’t.

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