TITLE: Inverses of elements in group algebras
QUESTION [6 upvotes]: If $G$ is a finite group whose elements are $g_1,\ldots,g_n$ and let $F$ be the field of real numbers $\mathbb{R}$ or the field of complex numbers $\mathbb{C}$. We define a vector space over $F$ with $g_1,\ldots,g_n$ as a basis, and we call this vector space $FG$, and all the elements in $FG$ have the form $$\lambda_1g_1 + \cdots  + \lambda_ng_n \quad(\text{all }\lambda_i \in F)$$
Now the vector space $FG$ is the group algebra of $G$ over $F$. And it also known that the identity element under multiplication in $FG$ is $1e$ where $1$ is the identity in $F$ and $e$ is the identity element in $G$.

My question is: given an element in the group algebra $FG$, can we
  find the inverse of that element (construct it explicitly)?

For example, let $$G = C_3 = \bigl\langle a: a^3 = 1 \bigr\rangle$$ be the cyclic group of order $3$, and let $F = \mathbb{R}$. Now let $u = a + a^2 \in FG$. Now I want to know what is the inverse of $u$, meaning that I want to find an element $v \in FG$ such that $uv =e =vu$. More formally, we need to find $$v= \lambda_1e + \lambda_2a + \lambda_3a^2$$ such that $uv=e=vu$, so I just multiply $$\begin{align*}
uv &= (a + a^2)(\lambda_1e + \lambda_2a + \lambda_3a^2) \\
&= \lambda_1a + \lambda_2a^2 + \lambda_3e + \lambda_1a^2 + \lambda_2e + \lambda_3a \\
&= \lambda_1(a + a^2) + \lambda_2(a^2 +e) + \lambda_3(e + a)
\end{align*}$$
But now what is the next step, I need $e + e + e = e$ but how can I achieve that ?

REPLY [3 votes]: Note that not every element has an inverse, as some are zero divisors. This element, however, does.
Note that 
$$(a+a^2)^2=a^2+2a^3+a^4=a+a^2+2e$$
so
$$(a+a^2)^2-(a+a^2)=2e$$
so
$$(a+a^2)(1-(a+a^2))=2e$$
hence
$$(a+a^2)\left(\frac{1}{2}(1-(a+a^2))\right)=e$$
so the inverse is
$$\frac{1}{2}(1-(a+a^2))$$
In general, left multiplication of an element $y=\sum_{g\in G}{a_gg}$ by an element $x=\sum_{h\in G}{b_hg}$ with unknown coefficients and setting it equal to $1e$ yields a system of equations, which we can solve for $b_h$ if the element is invertible. We have
$$\sum_{gh=g'}{a_gb_hg'}=1e$$
so we need to solve
$$\sum_{gh=g'\neq e}{a_gb_h}=0$$
and
$$\sum_{gh=e}{a_gb_h}=1$$
If this system has a solution, then $\sum_h{b_hh}$ is a left inverse for $y$. By symmetry of the roles of $g$ and $h$ in the summation, the element will also be a right inverse.