Find the upper central series of $Q_{2^n}$.

Solution 1:

It is not difficult to see from the presentation that the only nontrivial element in $Z := Z(Q_{2^n})$ is $x^{2^{n-2}}$, so $Z = \langle x^{2^{n-2}}\rangle$ has order $2$ and $$Q_{2^{n-1}}/Z = \left\langle x,y \mid x^{2^{n-2}} = y^2 = 1, y^{-1}xy = x^{-1} \right\rangle,$$ which is the dihedral group $D_{2^{n-1}}$.

So this reduces the problem to finding the upper central series of $$D_{2^n} = \left\langle x,y \mid x^{2^{n-1}} = y^2 = 1, y^{-1}xy = x^{-1} \right\rangle.$$ We claim that this is $$1 < \langle x^{2^{n-2}}\rangle <\langle x^{2^{n-3}}\rangle < \cdots \langle x^2 \rangle < D_{2^n}.$$ This is true for $n=1$ ($D_4 = C_2 \times C_2$), and we can prove it by induction by observing that $Z(D_{2^n}) = \langle x^{2^{n-2}} \rangle$, and $D_{2^n}/Z(D_{2^n}) \cong D_{2^{n-1}}$