The $i^{th}$ term in the upper central series of the dihedral group of order $2^n$ is equal to the $(n-1-i)^{th}$ term in its lower central series. The Problem: Prove that $Z_i(D_{2^n})=D_{2^n}^{n-1-i}$, where $D_{2^n}$ is the dihedral group of order $2^n$.
$Z_i(G)$ is the $i^{th}$ term in the upper central series of $G$, which is inductively defined by:$Z_0(G)=1$, $Z_1(G)=Z(G)$, and $Z_{i+1}(G)$ is the subgroup of $G$ containing $Z_i(G)$ such that $Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))$.
$G^i$ is the $i^{th}$ term in the lower central series of $G$, which is inductively defined by: $G^0=G$, $G^1=[G, G]$, $G^{i+1}=[G, G^i]$, where $[H, K]=\langle[h, k]|h\in H, k\in K\rangle$.
Source: Abstract Algebra, $3^{rd}$ edition by Dummit and Foote.
My Attempt: Induction on $i$. Let $G=D_{2^n}$.
Base case: $i=0$. Trivial by the fact that $D_{2^n}$ is nilpotent of class $n-1$ and the theorem that if $G$ is nilpotent of class $c$, then $G^{c-i}\leq Z_i(G)$ for all $i\in\{0, 1, \dots, c\}$.
Induction Hypothesis: Suppose $Z_i(G)=G^{n-1-i}$. We need to show that $G^{n-2-i}=Z_{i+1}(G)$.
Now, we know that $G^{n-2-i}\leq Z_{i+1}(G)$-i.e., $G^{n-2-i}\subseteq Z_{i+1}(G)$-by the theorem used in the base case. Thus it suffices to show that $Z_{i+1}(G)\subseteq G^{n-2-i}$.
Note $Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))$, where the RHS is abelian; thus $[Z_{i+1}(G), Z_{i+1}(G)]\leq Z_i(G)$ by a theorem about commutator groups. Hence $[Z_{i+1}(G), Z_{i+1}(G)]\leq G^{n-1-i}$ by the induction hypothesis, i.e., $[Z_{i+1}(G), Z_{i+1}(G)]\leq [G, G^{n-2-i}]$ (by the definition of $G^{n-1-i}$). Screeching halt.
Any hint would be greatly appreciated.
 A: Although you are following Dummit and Foote's numbering for the lower central series, I will note that this is very much not standard. The use of exponents rather than other notation is uncommon as well, since it clashes with the common notation $G^k$ for the subgroup generated by the $k$th powers of elements of $G$.
From my bookshelf:

*

*Susan McKay's Finite $p$-groups and Leedham-Green/McKay's Structure of Groups of Prime Power Order use $\gamma_i(G)$ to denote the $i$th term of the lower central series, and define $\gamma_1(G)=G$, $\gamma_{n+1}(G) = [\gamma_n(G),\gamma(g)]$.


*Marshal Hall's Group Theory uses $\Gamma_n(G)$, and numbers them as above.


*Isaac's Finite Group Theory uses exponents, but defines $G^1 = G$ and $G^{n+1}=[G^n,G]$.


*Rotman's Introduction to the Theory of Groups, 4th Edition, uses $\gamma_i(G)$, and defines $\gamma_1(G)=G$.


*Derek Robinson's A Course in the Theory of Groups, 2nd Edition, uses the same notation and numbering as Rotman, McKay, and Leedham.


*Hanna Neumann's Varieties of Groups defines $G_{(1)}=G$ and $G_{(n+1)}=[G_{(n)},G]$.
I will follow the usual notation; we just prove that the upper and lower central series of $D_{2^{n+1}}$ coincide.

It is easier, I think, to work with induction on $n$ and prove it all in a stroke. Recall that we have already shown that:


*

*$D_{2^{n+1}}/Z(D_{2^{n+1}}) \cong D_{2^n}$.

*$Z_{k}(G/Z(G)) = Z_{k+1}(G)/Z(G)$.

We add the following:
Lemma. If $f\colon G\to K$ is a surjective group homomorphism, then $f(G_n) = K_n$, where $F_n$ is the $n$th term of the lower central series of $F$.
Proof. The result holds for $n=1$, since $f$ is surjective. Assume that $f(G_n) = K_n$. If $x\in G_n$ and $g\in G$, then $f(x)\in K_n$, $f(g)\in K$, so $f([x,g]) = [f(x),f(g)]\in [K_n,K]=K_{n+1}$, hence $f(G_{n+1})\subseteq K_{n+1}$. Conversely, if $y\in K_n$ and $z\in K$, then there exists $a\in G_n$ and $g\in G$ such that $f(a)=y$ and $f(g)=z$. Then $[y,z] = f([a,g])\in f(G_{n+1})$; since all generators of $K_{n+1}$ lies in $f(G_{n+1})$, we conclude that $K_{n+1}\subseteq G_{n+1}$, giving equality. $\Box$
We do induction on $n$. We denote $D_{2^k}$ by
$$D_{2^k} = \langle r,s\mid r^{2^{k-1}}=s^2=1, sr=r^{-1}s\rangle.$$
For $n=3$, we are working with the dihedral group of order $8$. The upper central series is $Z_0(D_8) = \{e\}$, $Z_1(D_8) = \langle r^2\rangle$, $Z_2(D_8) = D_8$. The lower central series is $(D_8)_1 = D_8=Z_2(D_8)$, $(D_8)_2=\langle r^2=[r,s]\rangle=Z_1(D_8)$, $(D_8)_3 = \{e\}=Z_0(D_8)$; so the two series coincide.
Assume the result holds for $D_{2^n}$, and let $G=D_{2^{n+1}}$. Then by looking at $D_{2^{n+1}}/Z(D_{2^{n+1}})\cong D_{2^n}$, we know that $G_1=Z_{n}(G)$, $G_2=Z_{n-1}(G)$, and so on, by lifting the upper and lower central series from the quotient. Then we just need to note that $G_{n-1}=Z_2(G)$ is cyclic, not central, and has order $4$, so $G_{n}$ must be equal to its unique subgroup of order $2$, namely $Z_1(G)$ (cannot be trivial, cannot be itself). This proves the two series for $D_{2^{n+1}}$ also coincide.
