Normal Abelian Subgroup does not imply Abelian Quotient Group I'm a bit confused and just need some clarification about what I am missing in this:
I have $S_4$ with normal subgroup $N=\lbrace(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\rbrace$. 
I know that $N$ is normal (and abelian), which means $gN=Ng, \forall g\in G$, so to me, by definition of a quotient group it follows that $\frac{G}{N}=\lbrace gN:g\in G\rbrace=\lbrace Ng:g\in G\rbrace$, which suggests that $\frac{G}{N}$ is abelian, however I know this is wrong because $\frac{G}{N}\cong D_6$ which isn't abelian. 
So just wondering what step I am misunderstanding, thanks.

Recall that the group operation on $\frac{G}{N}$ is $(g_1N)(g_2N) = g_1g_2N$ if you use left cosets or $(Ng_1)(Ng_2) = Ng_1g_2$ if you use right cosets. Knowing that $gN = Ng$ does not imply $g_1g_2N = g_2g_1N$ but rather that $g_1g_2N = Ng_1g_2$.
