$Q$ group in form of $2\times2$ matrices closed for multiplication and inverses explanation If I define quarternion group as $Q=<A,B>=${ $I,-I,A,-A,B,-B,AB,-AB$ }, where 
$A= \begin{bmatrix}
  i & 0 \\
  0 & -i \\
 \end{bmatrix} B= \begin{bmatrix}
  0 & 1 \\
  -1 & 0 \\
 \end{bmatrix}$
is there an explanation why $Q$ is closed for multiplication and inverses that does not require using multiplication table or Cayley graph?
 A: It already says we have $\pm I, \pm A, \pm B$ and $\pm AB$ in there. There are three products such that, if $Q$ was not closed, at least one of them has to take us out of the set: $A^2, B^2$ and $BA$. If those products are in $Q$, we will know that any element of $Q$, when multiplied from any side by $A$ or $B$, will still be in $Q$, which is enough.
For instance $(AB)(AB) = A(BA)B$, so if $BA$ turns out to already be an element of $Q$, then $A(BA)$ is an element of $Q$, which makes $(A(BA))B$ an element of $Q$, and we are done.
A: See the effect of A and B on a column vector [x,y], and interpret the result geometrically.(e.g A[x,y]= [ix,-iy], which is the complex conjugate of [x,y] rotated by an angle of $\pi /2$ counter-clockwise)
(I assume that you're asking for intuition. If not, you can simply check by usual verification by calculation.)
A: Another method is to check that $A^4=B^4=1$ and $AB=B^{-1}A$ and then use the fact that the group defined by the presentation $\langle x,y \mid x^4=y^4=1, xy=y^{-1}x \rangle$ has order $8$. This is probably overkill for this particular problem, but the technique can be used on very large finite groups of matrices.
