I was asked to prove that $\langle x,y|x^8 = 1 , x^4 = y^2 , xy = y^{-1} x\rangle$ defines a $2$-group of order at most $16$. It is well-known that the group $\langle x,y|x^8 = 1 , x^4 = y^2 , xy = y^{-1} x\rangle$ defines the quaternion group $Q_{16}$ which has order $16$. Thus since it is finite and $16=2^4$ it is clearly a $2$-group. Is that sufficient? Or am I somehow misunderstanding this question?

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    $\begingroup$ Presumably your hiding the entire answer in an "it is well known that" will not go over well with whoever will be judging the answer to your question. $\endgroup$ – Mees de Vries Nov 14 '18 at 11:46
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    $\begingroup$ Better idea is to show that every element of the group can be written in the form $y^ix^j$ with $0\le i\le1$ and $0\le j\le7$. $\endgroup$ – Gerry Myerson Nov 14 '18 at 11:54
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    $\begingroup$ How about I first work with relations and show that the order is at most 16 (Gerry's comment) and then by using that fact and von Dyck's Theorem I can deduce that it is exactly 16 (which still requires knowing that the presentation defines $Q_{16}$)? $\endgroup$ – amator2357 Nov 14 '18 at 11:58
  • $\begingroup$ Assuming a strictly harder-to-prove version of your result will not go down well. Instead, note that all elements have order a power of 2, and obtain some relations limiting their numbers. Alternatively, apply rule #1: any time you say anything of the form "it is well known that", have a proof of that claim in mind. Now, write down that proof,and you're done. $\endgroup$ – user3482749 Nov 14 '18 at 12:42
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    $\begingroup$ Your "it is well-known" assertion is false. It is not true that this presentation defines $Q_{16}$. It defines a different group of order $16$. The group defined here has centre $\langle x^2 \rangle$ of order $4$, whereas $Q_{16}$ has centre of order $2$. $\endgroup$ – Derek Holt Nov 14 '18 at 15:08

Note that, by using the first two relations in the presentation, we can rewrite the third relation as $y^{-1}xy = y^{-2}x = x^{-4}x = x^5$.

Also, putting $z=yx^2$, we have $z^2 = (yx^2)^2 = y^2(y^{-1}x^2y)x^2 = x^4x^{10}x^2 = 1$, and $z^{-1}xz = x^{-2}y^{-1}xyx^2 = x^5$.

So the group is isomorphic to the group defined by the presentation $$\langle x,z \mid x^8=z^2=1, z^{-1}xz=x^5\rangle,$$ which manifestly a semidirect product of $\langle x \rangle$ of order $8$ with $\langle z \rangle$ of order $2$.

  • $\begingroup$ Thank you for that Derek and thank you for pointing other things out. $\endgroup$ – amator2357 Nov 14 '18 at 16:40

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