Using only upper central series, find the degree of nilpotency of the dihedral group $G = D_{16}$. (Answer: $3$)
So I need to find the non-negative integer $c$ such that $Z^c(G) = G$, where $Z^c(G)$ is defined as $$Z^0(G) = 1, Z^1(G) = Z(G), Z^2(G) / Z^1(G) = Z(G / Z^1(G)), Z^3(G) / Z^2(G) = Z(G / Z^2(G)), \cdots$$
And the upper central series is $$Z^0(G) \leq Z^1(G) \leq Z^2(G) \leq \cdots$$
I can rule out $c = 1$ since $Z^1(G) = Z(G) \neq G$ because $G$ is not a commutative group. If $c = 0$, then there are no elements in $G$ besides the identity that commute with every other element in the group, but I believe this is false since a rotation eight times commutes with every other rotation or reflection operations.
So now I want to check if $Z^2(G) = G$. I know that $$Z^2(G) / Z^1(G) = Z(G / Z^1(G)) = Z^2(G) / Z(G) = Z(G / Z(G))$$
I Googled the center of $D_{16}$, and it came out to be $Z(G) \cong \Bbb Z_2$. So then I have $$Z^2(G) / \Bbb Z_2 = Z(G / \Bbb Z_2)$$. Since the index of $\Bbb Z_2$ is $8$, $G/\Bbb Z_2 \cong \Bbb Z_8$. So $$Z^2(G) / \Bbb Z_2 = Z(\Bbb Z_8)$$. The center $Z(\Bbb Z_8) = \Bbb Z_8$ since $\Bbb Z_8$ is a commutative group. So then $$Z^2(G) / \Bbb Z_2 = \Bbb Z_8$$.
I'm not sure what to do from here. However, this procedure seems fairly long and I was wondering if there's a faster method to compute the nilpotency degree using upper central series.
Hint: $D_{16}/Z(D_{16})$ is isomorphic to $D_8$ and $D_8$ has nilpotency class $2$.
Let $\alpha$ be the elements of order $8$ and $\beta$ be an involution not contained in $<\alpha>$. It is easy to see that $\alpha^\beta=\alpha ^{-1}$. That means that $Z(G)$ does not contains such involution. Hence $Z(G)\leq <\alpha>$. If $\alpha^i$ is the generater of $Z(G)$, $(\alpha^i)^\beta=(\alpha^i)^{-1}$. Thus, $Z(G)=<\alpha^4>$.
Let $\bar G=G/Z(G)$. Then $\bar G=\{\bar \alpha,\bar \beta|\bar \alpha^4= \bar \beta^2=1 \ and \ {\bar \alpha}^{\bar \beta}= {\bar \alpha}^{-1} \}$. Hence, $\bar G $ is isomorphic to $D_8$. I left you to observe that $D_8$ has nilpotency class $2$ as it is easy. Actually any nonabelian group of order $p^3$ has this property.