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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 :)

Shaun
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  • Another exercise from Robinson's book. If you really can't solve it, it is strange. If you can then what is the goal? Did you get a permission from Derek Robinson? – markvs Dec 31 '21 at 17:23
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    It is a graduate textbook, @markvs. The exercises can be, thus, quite challenging; that I'm stuck is not all that strange. Brief excerpts are permitted by the copyright license. – Shaun Dec 31 '21 at 17:26
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    Show that $Z := Z(Q^{2^n})$ has order $2$ and that $Q_{2^n}/Z \cong D_{2^{n-1}}$ (the dihedral group). That reduces the problem to finding the upper central series of $D_{2^n}$, which can be done by induction, because $D_{2^n}/Z(D_{2^n}) \cong D_{2^{n-1}}$. – Derek Holt Dec 31 '21 at 17:47
  • Thank you, @DerekHolt. Please would you write that as an answer? – Shaun Dec 31 '21 at 17:52
  • See the text by Keith Conrad. It contains everything you wanted to know about the generalized quaternion group but were afraid to ask. See, for example, Theorem 4.9 and its proof in the Appendix. Conrad refers to the book by Scott, W. R. Scott, “Group Theory,” Dover, New York, 1987. https://kconrad.math.uconn.edu/blurbs/grouptheory/genquat.pdf – markvs Dec 31 '21 at 18:52
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    @markvs What is your issue, exactly, with this question? – the_fox Dec 31 '21 at 19:08
  • @the_fox: This is one of a series of questions copying exercises from a book by Derek Robinson. My issue is that it looks like Robinson did not give permission to publish his questions. Also the text of Conrad contains the description of the center of $D_{2^n}$ and the factor over the center, hence the upper central series. This is not an issue with the question, though. It just shows that the answer is trivial and can be found on google. – markvs Dec 31 '21 at 19:13

1 Answers1

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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 $D_{2^n}/Z(D_{2^n}) \cong D_{2^{n-1}}$

Derek Holt
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