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Given $\rho$, an irreducible representation of a finite group $G$, prove $\rho(id)=I$ (identity matrix with appropriate number of rows and columns $d_{\rho}\times d_{\rho}$) and $\rho(s^{-1})=\rho(s)^{-1}$ for $s\in G$.

This may be a really basic question but I'm having trouble proving it explicitly. By definition, $\rho$ is a function $\rho:G\rightarrow GL_n(\mathbb{C})$ such that $\rho(st)=\rho(s)\rho(t)$ for all $s,t\in G$.

By definition, $\rho(s)I=\rho(s)=\rho(s\cdot id)=\rho(s)\rho(id)$, so $\rho(id)=I$. But how can I assure that it is $d_{\rho}\times d_{\rho}$?

Secondly, I'm not sure how to go about showing the $\rho(s^{-1})=\rho(s)^{-1}$

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  • $\begingroup$ An easier way would be to show $\rho$ is a group homomorphism from $G$ to $GL_{n}(\mathbb{C})$, and use the properties of group homomorphisms. $\endgroup$ – user458276 Mar 4 at 23:51
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A representation $(V, \rho)$ of a finite group $G$ is given by a group homomorphism $\rho\colon G \to \mathrm{Aut}_{\mathbb{C}} (V) $. In particular, $\rho$ maps the identity in $G$ to the identity in $\mathrm{Aut}(V)$ and preserves the inverse. This holds even if the representation is not irreducible, you only use group homomorphism properties.

Just to clarify the steps: $$\rho(e_G) = \rho(e_G e_G) = \rho(e_G)\rho(e_G), $$ and since $\rho(e_G)$ is invertibile you get $\rho(e_G) = \mathrm{Id}$. Now if $s$ is an element of $G$, then $\mathrm{Id} = \rho(ss^{-1}) = \rho(s)\rho(s^{-1})$, from which your claim follows.

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  • $\begingroup$ How do you show that the identity matrix is in fact $d_{\rho} \times d_{\rho}$? $\endgroup$ – Ya G Mar 5 at 0:31
  • $\begingroup$ That $d_{\rho}$ is just some natural number. What happens is that the identity in the automorphisms of $V$ will be some square matrix, but this is obvious. There is nothing to prove. $\endgroup$ – Gibbs Mar 5 at 8:54

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