Possible Duplicate:
Isomorphism between $I_G/I_G^2$ and $G/G'$
Let $G$ be a finite group. Let $I\unlhd\mathbb{Z}[G]$ be the augmentation ideal of the integral group ring $\mathbb{Z}[G]$.
I'm trying to prove that
$G/G' \cong I/I^2$ as abelian groups
I defined a map $\varphi:G\rightarrow I/I^2$ by $\varphi(g)=(g-1)+I^2$. I showed that $\varphi$ is a group epimorphism. What's left is to show that $\ker\varphi=G'$. Since $\varphi$ is a homomorphism into an abelian group, we get $G'\subset \ker\varphi$.
So my question is
How to prove that $\ker\varphi \subset G'$?
My attempt at this part was to take generators for $I^2$ of the form $(g_0-1)(h_0-1)$ and show that if $g-1$ is a sum of such generators, then $g$ is in $G'$. In the case that $g-1$ is just one such generator, I could prove that actually $g=1$. I then tried to proceed by induction, but failed to do so.