Date modified: Thu 2023-05-11 . 11:51 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Center of dihedral groups. Define the dihedral group $Dih_n$ to be the symmetry group of a regular $n$-gon (and of $2n$ elements). We will show that they have center $Z(Dih_n)$ isomorphic to either $\mathbb Z_2$ or trivial, and compute the quotient $Dih_n/Z(Dih_n)$. Note the center $Z(G)$ of a group $G$ is the set of elements that commutes with everything in $G$. In particular, if $n\ge4$ is even, then $Z(Dih_n) = \{1, r^{n/2}\}$ and $Dih_n / Z(Dih_n) \approx Dih_{n/2}$. --- Some reminders. The dihedral group $Dih_n=\langle r,f \rangle$ is generated by a rotation $r$ and a flip $f$, where they satisfy the following relations: $f^2=1$, $r^n=1$, and $rf = fr^{n-1}$. The last relation can be understood geometrically, and is equivalent to a more memorizable form $rfrf=1$. Another useful derived relation is then $r^k f = fr^{n-k}=fr^{-k}$. The $2n$ elements of $Dih_n$ are then $1,r,\ldots,r^{n-1},f,rf,\ldots,r^{n-1}f$. When $n=1$, we see $r=1$ and we just have $f^2=1$. So $Dih_1 =\{1,f\}\approx \mathbb Z_2$. When $n=2$, we have three nontrivial order two elements: $r,f,rf$ (as $rf=fr$). So $Dih_2\approx V_4$ the Klein-4 group. In these two cases, both $Dih_1$ and $Dih_2$ are abelian, so the center is just itself and the quotient is trivial. --- Geometrically, each $fr^k$ is a flip of the regular $n$-gon about some axis. When $n \ge 3$ is odd, consider two distinct flips $fr^a$ and $fr^b$. So $fr^a fr^b = r^{b-a}$ and $fr^b fr^a = r^{a-b}$, which are inverses of each other. But since in an odd $n$-gon (with $n\ge 3$), no nontrivial rotation is its own inverse, these elements are distinct. So none of the flips are in the center. And for a nontrivial rotation $r^k$, it would not commute with $f$, since $r^kf = fr^{-k}$, which is distinct from $fr^k$. This is because that would require $r^k = r^{-k}$, and we remarked already that no trivial rotation is its own inverse in an $n\ge3$ gon. Hence $Z(Dih_n) = 1$ when $n\ge3$ odd. When $n\ge 4$ is even. Consider the rotation $r^{n/2}$. We claim this element is in the center. The element $r^{n/2}$ commutes with every rotation. Now consider any flip $fr^k$, we have $r^{n/2}fr^k = f r^{n/2}r^k = f r^k r^{n/2}$, as famously $n-(n/2) = n/2$. So we deduce that $r^{n/2} \in Z(Dih_n)$ when $n$ is even. However, for a nontrivial rotatoin that is not of order 2, say $r^k$, it would not commute with $f$, as $r^k f = fr^{-k}$ and for $fr^{-k} = fr^k$ we require $r^k = r^{-k}$, namely it being order 2. Since $r^{n/2}$ is the only order 2 rotation, no other rotation can be in the center (except identity). Similarly, for any flip $fr^k$, it would not commute with $r$. Note $rfr^k = fr^{k-1}$, and to have it equal to $fr^kr = fr^{k+1}$ requires $r^{k+1} = r^{k-1}$ , or $r^2 = 1$. But when $n\ge 4$, $r^2\neq 1$, so no flip is in the center. Hence $Z(Dih_n) = \{1,r^{n/2}\}$ when $n\ge 4$ is even. --- It remains to compute the quotient by the center in the interesting cases, namely when $n\ge 4$ is even. Write $Z= Z(Dih_n)$. Note the coset $r^k Z = \{r^k , r^{k+n/2}\}$ glues two rotations together, so we just have $n/2$ many cosets of the form $r^kZ$, for $0\le k < n/2$ . And the reflections $fr^kZ = \{ fr^k , fr^{k+n/2}\}$ glues two reflections toghether, so we just have $n/2$ many cosets of the form $fr^kZ$, for $0\le k < n/2$. These $n$ cosets are generated by $rZ$ and $fZ$, where $(rZ)^{n/2} = rZ$, $(fZ)^2 = fZ$, and $(rZ)(fZ)(rZ)(fZ)=Z$. Note these are the same relations as in $Dih_{n/2}$, and that $|Dih_{n/2}|=n$, the same size of $Dih_n/Z(Dih_n)|$. So we see that $Dih_n/Z(Dih_n) \approx Dih_{n/2}$ ! Note, above we use the fact that if $G$ and $H$ are both finite groups, $G$ has generators $g_1,\ldots,g_k$ and $H$ has the same number of generators $h_1,\ldots,h_k$. If the group $G$ is determined by some generator relations in $g_i$, and these relations are also observed by $h_i$ with the obvious correspondance $g_i \mapsto h_i$, then $G\approx H$ provided that $|G| = |H|$. This is because in this case, $H$ is a quotient of $G$, so with $|G|=|H|$ implies that $G/1\approx H$. #group-theory