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3 minor omission

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) $FG$-module, 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$annihilates every simple$FG$-module. kG$-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 $j.$

2 typo

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 $FG.$ 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$is a simple (sometimes called irreducible)$FG$-module, then$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$annihilates every simple$FG$-module. Then you can use the fact that every finite dimensional$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$and$\lambda \in k.$Then it is relatively easy to see that$1_{G} + \frac{j}{\lambda}$is invertible, using the nilpotency of$j.$1 Since you say this is an undergraduate thesis, I will take a few steps back. The augmentation ideal$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$FG.$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$is a simple (sometimes called irreducible)$FG$-module, then$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$annihilates every simple$FG$-module. Then you can use the fact that every finite dimensional$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$and$\lambda \in k.$Then it is relatively easy to see that$1_{G} + \frac{j}{\lambda}$is invertible, using the nilpotency of$j.\$