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 $8$? 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? 
 A: 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 $x^2=y^2 = z$.
A: Hint: Try to convert your representation in to the following representation:
$Q_{8}= \langle a,b\mid a^4=e,a^2=b^2,aba=b \rangle$
This relation shows that every element of $Q_8$ is of the form:
($a^{s} b^{t}$), $0\leq s\leq 3$,$0\leq t \leq 1$
Clearly, $\left |Q_8  \right | \leq 8$.
Next, we consider a subgroup of $GL(2,\mathbb{C})$ generated by two matrices:
$A=\begin{pmatrix}
 i& 0\\ 
 0& -i
\end{pmatrix},B=\begin{pmatrix}
0 & 1\\ 
 -1& 0
\end{pmatrix}$
It is easy to see that $A^4=I,A^2=B^2,ABA=B$
So the correspondence $a\mapsto A,b\mapsto B$ defines a surjective homomorphism $Q_8 \to \left \langle  A,B\right \rangle$.
On the other hand, $\left \langle  A,B\right \rangle$ has $8$ elements,namely:
$\left \langle A,B \right \rangle=\left \{\pm  \begin{pmatrix}
 1& 0\\ 
 0& 1
\end{pmatrix} ,\pm \begin{pmatrix}
 i&0 \\ 
 0& -i
\end{pmatrix},\pm \begin{pmatrix}
 0& 1\\ 
 -1&0 
\end{pmatrix},\pm \begin{pmatrix}
0 &i \\ 
 i& 0
\end{pmatrix}\right \}$
So, $Q_8 \cong\left \langle A,B \right \rangle$ has order $8$
