Q:

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?

A:

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 $2$.

