Prove $\rho(id)=I$ and $\rho(s^{-1})=\rho(s)^{-1}$

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 $$\rho(s^{-1})=\rho(s)^{-1}$$

• 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. – user458276 Mar 4 at 23:51

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.
• How do you show that the identity matrix is in fact $d_{\rho} \times d_{\rho}$? – Ya G Mar 5 at 0:31
• 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. – Gibbs Mar 5 at 8:54