Can we map $\Bbb{Z}[G]$ back to $G^{\text{ab}}$ using a group homomorphism? Let $G$ be a finite group and $\Bbb{Z}[G]$ be the group ring. 
Define $f : \Bbb{Z}[G] \to G^{\text{ab}}$ by $f(\sum\limits_{g \in G} x_g g) = \prod\limits_{g \in G} \tilde{g}^{x_g}$.  Then $f$ is a group homomorphism, where $\tilde{g}$ is the coset of $g$ in $G$'s abelianization.  Let's check:
$$
f( x + y) = \prod_{g \in G} \tilde{g}^{x_g + y_g} = \prod_{g \in G} \tilde{g}^{x_g} \prod_{g \in G} \tilde{g}^{y_g}
$$
by the Abelian property of $G^{\text{ab}}$.
But is this map well-defined i.e. since the abelianization of a group is a quotient and all? 
I think so since if $x = \sum_{g \in G} x_g g = \sum_{g \in G} y_g g= y$ then $f(x) = \prod_{g \in G} \tilde{g}^{x_g} = f(y)$.  I think, but I'm not 100% sure yet.
If yes or no, I would also like to know if you can come up with some other homomorphisms $\Bbb{Z}[G] \to G$ or $G^{\text{ab}}$. 

Okay, so apparently it is indeed a group hom.  But I want now to compute its kernel.  I think it might have something to do with the norm element of the group ring, $N_G = \sum_{g \in G} g$.  
 A: The additive group of $\mathbb{Z}[G]$ is the free abelian group on the set underlying $G$. Since we have a quotient map $G \to G/[G, G]$, it follows that there is an induced homomorphism $\mathbb{Z}[G] \to G/[G, G]$. Well-definedness is only an issue for maps defined on quotients.
A: Consider the composition $\gamma:\mathbb{Z}[G]\to\mathbb{Z}[G^{\mathrm{ab}}]\to G^{\mathrm{ab}}$. Note $G^{\mathrm{ab}}=G/G'$ where $G'=[G,G]$.
The ring homomorphism $\mathbb{Z}[G]\to\mathbb{Z}[G/N]$ (where $N\trianglelefteq G$) has kernel $K=\bigoplus_{G/N} gI_N$, where
$$ I_N = \left\{\sum_{n\in N} a_n x_n\mid \sum_{n\in N} a_n=0\right\}=\bigoplus_{n\in N}\mathbb{Z}\cdot(n-e) $$
is the augmentation ideal (the kernel of the coefficient-summing map $\mathbb{Z}[N]\to\mathbb{Z}$).
And $\mathbb{Z}[G^{\mathrm{ab}}]$ (under addition) is the free abelian group on $G^{\mathrm{ab}}$'s underlying set, so we expect the kernel of the multiply-together map $m:\mathbb{Z}[G^{\mathrm{ab}}]\to G^{\mathrm{ab}}$ to be generated by $G^{\mathrm{ab}}$'s multiplication table, i.e.
$$ K'=\sum_{x,y\in G^{\mathrm{ab}}} \mathbb{Z}\cdot(x+y-xy). $$
Pulling $K'$ back to $\mathbb{Z}[G]$ yields the kernel of the composition $\gamma$. Pulling a $\mathbb{Z}$-basis for $K'$ back to a set in $\mathbb{Z}[G]$ and taking the union with a $\mathbb{Z}$-basis for $K$ should yield a $\mathbb{Z}$-basis for the kernel of $\gamma$.
