Assume the dihedral group of order $2n$ is defined by $$D_n=⟨s,r∣r^n=s^2=e,srs^{-1}=r^{-1}⟩$$ I know that $D_n$ is nilpotent iff $n=2^m$ for some $m\ge0$. I am asked to directly show that $G=D_4$ is nilpotent by calculating the ascending central series $\zeta^{i}(G)$ of it. I started by calculating:
$$\zeta^{0}(G)=\{e\}$$
$$\zeta^{1}(G)=Z(G)=⟨r^{2}⟩=\{e,r^2\}$$
The definition of $\zeta^{2}(G)$ is so that $\zeta^{2}(G)/\zeta^{1}(G)=Z(G/\zeta^{1}(G))$.
- How can one calculate $G/Z(G)$, without having to consider all the different cases "brute force"?
- How do you use this to find $\zeta^{2}(G)$? I am told to use the fourth isomorphism theorem, but I am not quite sure how.
Thanks in advance.
Best Answer
For dihedral groups of order $2^{n+1}$, $G/Z(G)$ is isomorphic to the dihedral group of order $2^n$.
(In my notational convention, $D_{2n}$ is the dihedral group of order $2n$, rather than of order $4n$; I also use $Z_k(G)$ to denote the $k$th center, instead of using $\zeta$).
Theorem. Let $n\gt 0$ and let $G$ be the dihedral group of order $2^{n+1}$, $$G = D_{2^{n+1}}\langle r,s\mid r^{2^n}=s^2=1,\quad sr=r^{-1}s\rangle.$$ Then $Z(G)=\langle r^{2^{n-1}}\rangle\cong C_2$, and $G/Z(G)$ is isomorphic to $D_{2^n}$, the dihedral group of order $2^n$.
Proof. If $r^is^j\in Z(G)$, $0\leq i\lt 2^n$, $0\leq j\lt 2$, then $r^{i+1}s^j = r(r^is^j) = (r^is^j)r = r^{i+(-1)^j}s^j$. Therefore, $(-1)^j=1$ so $j=0$. Then $r^{-i}s = s(r^i) = (r^i)s$, so $r^i=r^{-i}$, hence $2^{n}|2i$, so either $i=0$ or $i=2^{n-1}$. Thus, the only possible central elements are $1$ and $r^{2^{n-1}}$. These are central, which establishes the claim about $Z(G)$.
Let $H=G/Z(G)$. This is of order $2^n$; let $x$ be the image of $r$ and $y$ be the image of $s$. Since $s\notin Z(G)$, $y$ is of order $2$; since the smallest power of $r$ that lies in $Z(G)$ is $r^{2^{n-1}}$, $x$ is of order $2^{n-1}$. Finally, we have that $$yx = srZ(G) = r^{-1}sZ(G) = x^{-1}y.$$ By von Dyck's Theorem, there is a surjective homomorphism $D_{2^n}\to H$ mapping the rotation to $x$ and the reflection to $y$; since $D_{2^n}$ and $H$ have the same order, this surjection is an isomorphism, completing the proof. $\Box$
Corollary. The upper central series of $G=D_{2^{n+1}}$ is given by $Z_i(G) = \langle r^{2^{n-i}}\rangle$, $i=0,1,\ldots,n-2$, and $Z_n(G)=G$. In particular, $D_{2^{n+1}}$ is nilpotent of class exactly $n$, and so is a $2$-group of maximal class.
Proof. $Z_0(G)=\{e\} = \langle r^{2^n}\rangle$ and $Z_1(G)=\langle r^{2^{n-1}}\rangle$. Applying induction we get the result, up until we get to $D_4$, which is abelian and hence equal to its center. A $p$-group of maximal class is a $p$-group of order $p^k$ and nilpotency class $k-1$, so the final claim follows.
To use the 4th Isomorphism Theorem, note that by definition $\zeta^{k+1}(G)$ is the group whose image is the center of $G/\zeta^k(G)$. So you want to find the center of the quotient and lift it to the original group.