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Let $G$ be a group and $f,g: G \rightarrow GL(V)$ be two faithful representations over some field $K$ with $f:x\mapsto f(x)$ and $g:x \mapsto f(x^{-1})$.

I would like to find out if $f$ and $g$ are isomorphic. So I need an ismomorphism $\alpha: V \rightarrow V$ with $$\alpha \circ f(x) \circ \alpha^{-1} = f(x^{-1}). $$ What could such an isomorphism be? Does it depend on $K$?

Thank you for hints.


Later edit. As it is not possible to solve this in general: Is there an answer to the example $G=C_p$ cyclic group, $K$ some field of characteristic zero, $f,g$ irreducible representations of degree $> 1$?

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Are $f,g$ faithful? If not, I doubt it. –  Easy Apr 20 '13 at 14:49
    
yes, they are both faithful –  user73521 Apr 20 '13 at 14:55
    
Faithfulness has very little to do with this. Abelianness (module the kernel of $f$) of $G$, on the other hand, is sort of important. –  Mariano Suárez-Alvarez Apr 20 '13 at 16:43
    
I don't understand the question. Is it how to decide when $f$ and $g$ are isomorphic? To classify the set of representations $f$ for which $f(x^{-1})$ is an isomorphic representation? –  S123 Apr 20 '13 at 17:09
    
I would like to decide if $f$ and $g$ are isomorphic or not. The only information I have is that $g(x)=f(x^{-1}). Do you mean, that both cases can occur and this depends on the group, the field,... ? –  user73521 Apr 20 '13 at 17:16

2 Answers 2

up vote 3 down vote accepted

It depends. There are two separate issues here. Firstly, if $f$ is a faithful representation such that $g(x)=f(x^{-1})=f(x)^{-1}$ is also a representation, then the group $G$ is necessarily abelian:

$$f(x)^{-1}f(y)^{-1}=g(x)g(y)=g(xy)=f(xy)^{-1}=(f(x) f(y))^{-1}$$ implies that $f(x)$ and $f(y)$ commute and hence that $G$ is abelian.

Secondly, some representations (for instance, the inclusion of $\{\pm 1 \}$ in the multiplicative group of your field) have the property that $f$ and $g$ are isomorphic and others do not---for instance, the defining representation of the group of third roots of unity (working over the complex numbers) is not isomorphic to its dual $g$.

In response to the edited question, calculate the character. When working with subfields of the complex numbers, the representation will be self-dual precisely when the character is real-valued.

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Thank you for your help –  user73521 Apr 20 '13 at 17:29
    
You're welcome! –  S123 Apr 20 '13 at 17:32

If $f$ is faithful and $g$ is a homomorphism of groups, then $G$ is abelian. It follows that $V$ is a direct sum of representations of degree $1$, and it is easy to see that it is enough to suppose that $V$ is of degree $1$, for the general case follows from this.

Can you do that case?

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The problem here is, that the field must not be $\mathbb{C}$, so that the representations can have degrees other than 1. –  user73521 Apr 20 '13 at 17:03
    
The point of this answer is that, using its observations, you can easily construct examples where the tworepresentations are not isomorphic. –  Mariano Suárez-Alvarez Apr 20 '13 at 17:11
    
If I try to calculate e.j. $G=C_p$ cyclic, $K$ some field of characteristic zero and $f,g$ have degree bigger than one and be irreducible. Are they always not isomorphic then, or do I need to know how $f$ and $g$ looks exactly? –  user73521 Apr 20 '13 at 17:21
    
Well, using the observations of my answer you can easily construct $f$s which are isomorphic to the corresponding $g$s, so they are not always not isomorphic! –  Mariano Suárez-Alvarez Apr 20 '13 at 17:23
    
Thanks for helping –  user73521 Apr 20 '13 at 17:33

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