Originally Posted by

**Deveno** but φ doesn't have to be onto. look, say i have this example:

$\displaystyle \phi:S_3 \to \mathbb{Z}_2 \times \mathbb{Z}_2$ given $\displaystyle \phi((1 2 3)) = (0,0), \phi((1 2)) = (1,0)$.

that is a perfectly good homomorphism, and it is NOT surjective.

now explicitly, G' = {e, (1 2 3), (1 3 2)}, as one can see by computing all 36 commutators. so in this case, we have G/G' = {G', (1 2)G'}.

so the map $\displaystyle \widehat{\phi}$ is given by:

$\displaystyle \widehat{\phi}(G') = (0,0), \widehat{\phi}((1 2)G') = (1,0)$ which is not surjective.