$\displaystyle G-N$ is abelian since every element is it's own inverse (I proved this in the last homework I had).Let $\displaystyle G=S_4$ be the group of all permutations of $\displaystyle \{1,2,3,4 \}$.

The subset $\displaystyle N=\{1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$ of $\displaystyle G$ is a normal subgroup of $\displaystyle G$.

By proposition 3.5 of the notes, since $\displaystyle |G-N|=|G|$/$\displaystyle |N|=6$,either $\displaystyle G -N \cong C_6$ (a cyclic group of order 6) or $\displaystyle G-N \cong D_6$.

Decide which and explain your answer.

$\displaystyle D_6$ is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

For two groups to be isomorphic, they must be structurally similar (ie. $\displaystyle G-N$ must be isomorphic to another abelian group).

Therefore $\displaystyle G-N \cong C_6$.

I think this is right, but one of my friends has got it isomorphic to $\displaystyle D_6$. Which one of us is right? (I can't see a flaw in my reasoning).