I know the subgoups of the quaternion group (and that they are all normal), but i'm not sure how to determine the factor group Q8/N for each of the normal subgroups N.
Thanks for your help.
if $\displaystyle |N| = 1, 4$ or $\displaystyle 8$, then $\displaystyle Q_8/N \cong Q_8, \ C_2,$ or $\displaystyle \{1 \}$ respectively. the only (normal) subgroup of order 2 is $\displaystyle N=\{1,-1\}.$ then $\displaystyle Q_8/N = \{\bar{1}, \bar{i}, \bar{j}, \bar{k} \} \cong C_2 \times C_2.$ (here $\displaystyle C_2$ is the cyclic group of order 2.)
you probably know that $\displaystyle V=C_2 \times C_2$ is also called the Klein 4-group.
Q_8/{1,-1} is NOT {1,i,j,k} but the images of these elements in the factor group, just Noncommalg denoted by putting a bar over the elements i,j,k.
Why is this thing isomorphic with Z_2 x Z_2? Well, check that any of the elements 1, i, j, k squared gives you an element in {1,-1}...
Tonio