The converse: If $G$ is abelian and if $N\le G$, prove that $G/N$ is abelian

Question

Here is a problem from Herstein's "Topics in Algebra":

If $G$ is abelian and if $N$ is any subgroup of $G$, prove that $G/N$ is abelian

I have already proved this; it's quite simple. There's a very natural question that arises: is the converse true?

I say it's not. The reason is because the quotient group $G/N$ always require $N$ to be normal in $G$, and although every subgroup of an abelian group is normal the converse is not true. So $G$ is not abelian.

Is my intuition faulty?

Answer 1

Take $S_3$- it's a group of order 6, hence by Lagrange, any subgroup is of prime order, hence cyclic and abelian. However' $S_3$ isn't abelian