# 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.

Edit: Also, as far as I'm aware, what you call the left regular representation is called the left regular action. It may just be two ways of saying the same thing, but as you had (??) in your post I thought I'd add this in anyway.

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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