# Isomorphism of faithful representations

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 $g(x)=f(x^{-1}). Do you mean, that both cases can occur and this depends on the group, the field,... ? – user73521 Apr 20 '13 at 17:16 ## 2 Answers It depends. There are two separate issues here. Firstly, if$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. - 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? - 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