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Let $G$ be a finite group, $K$ a field which characteristic does not divide the group order and $V$ a $K$ vector space. Suppose there is an irreducible representation $f: G \rightarrow GL(V)$, $x \mapsto f(x)$. What can one say about a representation like $g: x \mapsto f(x^{-1})$ , $x\in G$?

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Unless $G'\leq \rm{ker}(f)$, that is not a representation of $G$. – Tobias Kildetoft Feb 11 '13 at 18:04
$f$ and $g$ are both assumed as representations of $G$. What do you mean with "$G'$"? – user61052 Feb 11 '13 at 18:07
By $G'$ I mean the derived (ie, commutator) subgroup of $G$. Unless all elements in the image of $f$ commute, then defining a new map by inverting the images of each element does not give a homomorphism. – Tobias Kildetoft Feb 11 '13 at 18:08
So if you know that $f$ and $g$ are both representations, then $f$ is an irreducible representation of the abelian group $G/G$, and thus $f$ is linear. – Andreas Caranti Feb 11 '13 at 18:12
Does $g$ also have to be irreducible? – user61052 Feb 11 '13 at 18:12
up vote 1 down vote accepted

A representation is a homomorphism, so if $g:x\mapsto f(x^{-1})$ is a homomorphism, then $g(x)g(y)=g(xy)$ so $f(x^{-1})f(y^{-1})=f(y^{-1}x^{-1})$ which, since $f$ is a homomorphism, is equal to $f(x^{-1}y^{-1})$. Thus $1=f([x,y])$ for any $x,y\in G$. Thus $G'\leqslant \text{ker}(f)$, where $G'$ is the derived subgroup $G'=\langle [x,y]:x,y\in G\rangle$. So, the image of $f$ in $GL(V)$ must be abelian.

Since $G'\leqslant \text{ker}(f)$, $f$ must be lifted from an irreducible representation of $G/G'$. If $K$ is algebaically closed, then every irreducible character of an abelian group is linear, so since $G/G'$ is abelian we must have that $f$ has dimension $1$.

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