# Presentation of the quaternion group $Q_{16}$

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?

• 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. – Mees de Vries Nov 14 '18 at 11:46
• 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 $0\le j\le7$. – Gerry Myerson Nov 14 '18 at 11:54
• 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 $Q_{16}$)? – amator2357 Nov 14 '18 at 11:58
• 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. – user3482749 Nov 14 '18 at 12:42
• 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 $2$. – 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 $$2$$.