Let $A$ be a frobenius complement for a group $G$ i.e. $A$ act on $G$ by automorphism s.t. $C\_A(g)=e$ for all nonidentity $g$.


Now, Action of $A$ can be linearly extended so that $A$ act on $F[G]$. As a result we can see $F[G]$ as a $F[A]$ module. Let $\chi$ be corresponding character of $A$ then 


$$\ \chi(g) = \begin{cases} 
 |G| & \textrm{ if $g=e$ } \\
 1 & \textrm{ if $g\neq e$ } \\
 \end{cases} \ $$
as one can easily compute.


Now, I am asking the converse of this situation; if $A$ has a character $\chi$ s.t.


$$\ \chi(g) = \begin{cases} 
 n & \textrm{ if $g=e$ } \\
 1 & \textrm{ if $g\neq e$ } \\
 \end{cases} \ $$


then can we say that $A$ is a frobenius complement for a group of order $n$ ?


**Note:** $F$ can be taken as $\mathbb C$ complex field and I have asked it [there](https://math.stackexchange.com/questions/1036514/can-we-say-that-a-is-a-complement-for-a-group-g) but it think it is suitable for here.



Can we say that $A$ is a complement for a group $G$?
As is so often the case, the answer is "not in general". Take $n=1+pq,$ where $p,q$ are primes with $q|p−1$. Let $A$ be a non-Abelian group of order $pq.$ Then $A$ is not a Frobenius complement, but $A$ does have a (complex) character which takes value $n$ on the identity and $1$ everywhere else. The character in question is the sum of the trivial character and the regular character. 


