# left regular representation of $G$ with respect to $g$

In the lecture, the prof used the fact that the ($\lambda_g$) left regular representation(??) is a bijection to prove cayley's theorem.

Definition: left regular representation of $G$ with respect to $g$ $$\lambda_g:G\longrightarrow G:h\longmapsto gh$$

Why is this true and how can I prove this?

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 Are you learning about group actions? If so you should look at this question: math.stackexchange.com/q/245038/12952 – Alexander Gruber♦ Dec 9 '12 at 19:32

A function is a bijection iff it has an inverse, so one way to prove the statement is to show that the inverse function to $\lambda_g$, $(\lambda_g)^{-1} = \lambda_{g^{-1}}$.
You could also just prove that $\lambda_g$ is surjective and injective directly, which isn't too hard to do.
To show that $\lambda_g$ is surjective, take $h \in G$. Then $\lambda_g(g^{-1}h) = gg^{-1}h = h$. $g^{-1}h \in G$ so we're done.
To show that $\lambda_g$ is injective, suppose $\lambda_g(h) = \lambda_g(h')$. Then $gh = gh'$ so by multiplying both sides on the left by $g^{-1}$ we have $h = h'$, so $\lambda_g$ is injective.
 Thank you for the answer. I want to prove that $\lambda_g$ is surjective and injective directly (I know the definitions), but how? – Onur Dec 9 '12 at 20:08 @OC89 I'll update the answer above to show how to do it, hang on just a minute! – Tom Oldfield Dec 9 '12 at 20:10 Ok, thank you ! – Onur Dec 9 '12 at 20:11 @OC89 done, hope that helps! – Tom Oldfield Dec 9 '12 at 20:14