TITLE: Is there a way to show $\langle x,y,z: xz=zx,yz=zy,xy=yxz,x^4=y^4=z^2=1\rangle$ has order $8$?
QUESTION [6 upvotes]: The quaternion group has a particular presentation
$$
\langle x,y,z: xz=zx,yz=zy,xy=yxz,x^4=y^4=z^2=1\rangle
$$
So it must have order $8$, but can you deduce that just from the relations? The relations show that anything can be put in form $x^iy^jz^k$ for $0\leq i,j\leq 3$, and $0\leq k\leq 1$, so there are at most $32$ elements. 
How can you eliminate redundancies?

REPLY [2 votes]: You have already shown that your group
$$
P = \langle x,y,z: xz=zx, yz=zy, xy=yxz, x^4=y^4=z^2=1 \rangle
$$
has order at most  $32$. To prove it has exactly order $32$, one way is to construct a group $G$ of order $32$ and elements $X, Y, Z$ in it which satisfy the relations for $x, y, z$. So $G$ is a homomorphic image of $P$, and $P$ has order at least $32$.
One such $G$ is the semidirect product of the abelian group 
$$
\langle Y \rangle \times \langle Z \rangle,
$$
with $X$ of order $4$ and $Z$ of order $4$, by a cyclic group $\langle X \rangle$ of order $4$, acting via the automorphism (of order $2$)
$$
Y^X = Y Z, \qquad Z^X = Z.
$$
Now note that the relations you gave to define $P$ are certainly satisfied in $Q_{8}$, taking $x = i$, $y = j$ and $z = -1$. But they do not define $Q_{8}$, as we have just seen. To do that you have to modify them, for instance as
$$
P = \langle x,y,z: xy=yxz, x^2=y^2 = z, z^2=1 \rangle,
$$
where I have omitted $xz=zx, yz=zy$, which now follow from $x^2=y^2 = z$.