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.

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- Oct 8th 2009, 04:43 AMLouiseFactor groups for the quaternion group?
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. - Oct 8th 2009, 05:08 AMNonCommAlg
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. - Oct 9th 2009, 01:25 AMLouise
How do you do the calculation Q8/{1,-1} to give {1,i,j,k}?

- Oct 9th 2009, 01:29 AMtonio
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