What does this linear system look like? Let $G = \{g_1, \dots, g_n\}$ be a finite abelian group and define $F[G]$ to be the set of all linear combinations of elements of $G$ with coefficients in the field $F$.  Then it forms a ring when we allow distributivity in the natural way.  For it to form a field, let $a = a_1 g_1 + \cdots + a_n g_n$, we need $b = b_1 g_1 + \cdots + b_n g_n$ s.t. $ab = 1$.  That is,
$$
ab = (a_1 g_1 + \cdots +a_n g_n)(b_1 g_1 + \cdots + b_n g_n) = \sum_{k=1}^n (\sum_{g_i g_j = g_k} a_i b_j)g_k = 1g_1 + 0 + \cdots + 0.
$$
So there's a linear system involved.  Does this linear system always have a unique solution?
 A: In general it won't have any solution. For example, if $a=\sum_{g\in G}g$ then $ag=a\Rightarrow a(g-1)=0$ for all $g\in G$ (by symmetry), so $a$ is a zero divisor hence has no multiplicative inverse.
A key idea here is that the elements of $G$ have no linear relations in $F[G]$, whereas units in $K^\times$ for a field $K$ will pretty much always have linear relations, like $1+(-1)=0$ or $1+\omega+\omega^2=0$. It is for this reason that elements of $G$ in $F[G]$ do not really look like units in a field.
In any commutative monoid $M$ (a set with associative multiplication, for example a ring including its $0$ element under multiplication) if $ab=e$ for fixed $a$ has a solution in $b$ then that solution is unique. For suppose $b,b'$ are two solutions. Then $b=eb=(ab')b=(ab)b'=eb'=b'$. Therefore for this setting, if there is a solution to $ab=1$ in $F[G]$ (for $b$; fixed $a$) then it must be unique.
For the general structure theory of finite-dimensional abelian group algebras, it pays to make use of the fact $F[A\times B]\cong F[A]\otimes F[B]$, incorporate the structure theory of finite abelian groups (either invariant factor or primary decompositions), and the fact that $F[C_n]\cong F[X]/(X^n-1)$ will by isomorphic to a direct sum of cyclotomic extensions of $F$ determined be the arithmetic relationship between $F$'s roots of unity and $n$ (apply Chinese Remainder Theorem and factor $X^n-1$).
