an example of a non-abelian group $G$ containing a proper normal subgroup $N$ such that $G/N$ is abelian. I am wondering if my claim is correct, and how to formalize/write it? Thank you.

Give an example of a non-abelian group $G$ containing a proper normal subgroup $N$ such that $G/N$ is abelian.

So I am thinking of the dihedral group of triangle, with $3$ rotations and $2$ reflections. This is not abelian; but if I quotient $3$ rotations, the group has only two elements, which must be abelian.
 A: Perhaps the cheapest sort of example would be the direct product of any non-abelian group with any non-trivial abelian group.
A: Yes, your example is fine. It is a special case of the fact that
$$
S_n/A_n \cong (\{\pm 1\},\times)
$$
where $n=3$.
More generally, you are looking for a surjective homomorphism from $G$ to an abelian group. By the first isomorphism theorem, $G$ modulo the kernel will be abelian. (And so the kernel must be non-trivial if $G$ is non-abelian)
A: One of the typical examples is the free group $F_2=\langle a\rangle*\langle b\rangle$ on two generators $a$ and $b$. There is a unique homomorphism $f:F_2\to\Bbb Z\times\Bbb Z$ sending $a$ to $(1,0)$ and $b$ to $(0,1)$. The kernel of $f$ is a normal subgroup $N$, and by the first isomorphism theorem we have an isomorphism $F_2/N\cong\Bbb Z^2.$
A: Your idea is just fine. 
Remember that all groups of order five or less are Abelian. This means that any not simple, not Abelian group of order 10 or less is an example. (Actually the smallest not Abelian simple group has order 60, so we're good for all not Abelian groups with order 10 or less :) )
Hint for a different example: Any nontrivial subgroup of the quaternion group.
A: One example would be $D_{2n}$:
$$D_{2n}=\mathbb Z_n\rtimes\mathbb Z_2$$
If $D_{2n}=\langle a,x\rangle$ where $\mathbb Z_n=\langle a\rangle\unlhd D_{2n}$ and $\langle x\rangle=\mathbb Z_2$ then $D_{2n}$ is an extention of $\langle a\rangle$ by $\mathbb Z_2$.
And another one could be 
$$T=\langle a,b\mid a^6=1, a^3=b^2=(ab)^2\rangle=\mathbb Z_3\rtimes\mathbb Z_4$$ of order $12$.
A: All non-abelian groups $G$ of order $\leq 10$ with a non-trivial center subgroup $Z$ possess this property.
The center subgroup is a normal subgroup of $G$.  It is a proper subgroup since $G$ is non-abelian.  Thus, since $G$ has at most $10$ elements and $Z$ is non-trivial, $G/Z$ has at most $5$ elements, and thus is abelian (the smallest non-abelian group has order $6$).
This includes, for example, the quaternion group, which has a center of cardinality $2$.
