This is part of Exercise 5.3.3 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
The Details:
(This can be skipped.)
On page 125 of the book,
[T]he upper central series
$$1=\zeta_0 G\le \zeta_1 G\le \zeta_2 G\le\dots,$$
defined by $\zeta_{n+1}G/\zeta_nG$ [as] the centre of $G/\zeta_nG$. [. . .] Of course $\zeta_1G=\zeta G$.
Here $\zeta G=Z(G)$, the centre of $G$.
On pages 140 to 141 of the book,
An important type of finite $2$-group that occurs in many investigations is the generalised quaternion group $Q_{2^n}$, ($n\ge 3$); this is a group with a presentation of the form
$$\left\langle x,y\ \middle|\, x^{2^{n-1}}=1, y^2=x^{2^{n-2}}, y^{-1}xy=x^{-1}\right\rangle.$$
I think the following theorem from the book, page 143, is relevant.
Theorem 5.3.6: A finite $p$-group has exactly one subgroup of order $p$ if and only if it is cyclic or a generalised quaternion group.
The Question:
Find the upper [. . .] central series of $Q_{2^n}$.
Thoughts:
If $n=3$, then $Q_{2^n}$ is the standard quaternion group of order eight. Its upper central series is
$$1\le \Bbb Z_2\le Q_8$$
according to Group Names. This can be verified using the First Isomorphism Theorem, a tool that I suspect will be helpful for $n\ge 4$.
For $n=4$, again on Group Names, the upper central series is
$$1\le \Bbb Z_2\le \Bbb Z_4\le Q_{16}.$$
For $n=5$,
$$1\le \Bbb Z_2\le \Bbb Z_4\le \Bbb Z_8\le Q_{32}.$$
For $n=6$,
$$1\le \Bbb Z_2\le \Bbb Z_4\le \Bbb Z_8\le \Bbb Z_{16}\le Q_{64}.$$
For $n=7$,
$$1\le \Bbb Z_2\le \Bbb Z_4\le \Bbb Z_8\le \Bbb Z_{16}\le \Bbb Z_{32}\le Q_{128}.$$
This suggests the pattern
$$1\le \Bbb Z_{2}\le\Bbb Z_{4}\le \dots\le\Bbb Z_{2^{n-2}}\le Q_{2^n}.$$
I don't see how to prove this. I think it has something to do with the relation $y^2=x^{2^{n-2}}$ in the presentation above.
As I said above, the First Isomorphism Theorem might help.
I haven't worked with upper central series before; presentations, on the other hand, are things I'm comfortable with.
Please help :)
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 $D_{2^n}/Z(D_{2^n}) \cong D_{2^{n-1}}$