Since you say this is an undergraduate thesis, I will take a few steps back. The augmentation ideal $I$ of the group algebra $kG$ is $\{ \sum_{g \in G} \alpha_{g}g: \sum_{g \in G} \alpha_g = 0\}.$ It is easy to see (in several ways) that $I$ is a two-sided ideal of $kG.$ One way is to note that it is the annihilator of the trivial module, which is $1$-dimensional, with a $k$-basis $\{v \}$ such that $vg = v $ for all $g \in G.$ No element of $I$ is a unit, as $I$ is a proper ideal. It remains to prove that every element of $kG \backslash I$ is invertible. Again, there are several ways to do this: one is to note that if $M$ is a simple (sometimes called irreducible) $kG$-module, then $G$ fixes a non-zero vector of $M$, so $M$ must be the trivial module. I leave this to you to do, or to research. Then it follows that $I$ annihilates every simple $kG$-module. Then you can use the fact that every finite dimensional $kG$-module has a composition series to see that $I^{n}$ annihilates the regular module $kG$ for some integer $n.$ It follows in particular that every element of $I$ is nilpotent.
Every element of $kG \backslash I$ is of the form $\lambda 1_G + j$ for some $j \in I$ and
nonzero $\lambda \in k.$ Then it is relatively easy to see that $1_{G} + \frac{j}{\lambda}$ is invertible, using the nilpotency of $j.$

doesbelong to MO. – Qfwfq Aug 27 '11 at 21:10