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I don't really understand the definition of left regular representation of $G$. Could anyone give me some explanation or examples?

The permutation representation afforded by left multiplication on the elements of $G$ (cosets of $H = 1$) is called the left regular representation of $G$.

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Did you try ? – Dietrich Burde Sep 22 '13 at 20:34
In the wiki article, they still use the term "afford". What does it mean? – Tumbleweed Sep 22 '13 at 20:38
Hmm, means that $L(x)$ sends $g$ to $xg$ (multiplication from the left). I don't know why they use "afford" ( – Dietrich Burde Sep 22 '13 at 20:46
Hmm~ Now I get it, thank you! @DietrichBurde – Tumbleweed Sep 22 '13 at 20:50
if you can see my answer for… it would be more clear i guess... left regular representation is a homomorphism of $G$ to $S_{|G|}$ considering multiplication of fixed element of $G$ by each element of $G$ giving an element in $S_{|G|}$... – Praphulla Koushik Sep 23 '13 at 6:18
up vote 2 down vote accepted

Let $G$ be a group, then you can get an homomorphisms $$ f \colon G \to S(G)$$ where $S(G)$ is the group of permutations on $G$, defined as the mapping that send every $g \in G$ in the mapping $f(g) \colon G \to G$ such that for every $x \in G$ the equation $$f(g)(x)=gx$$ holds.

Associativity and unit axioms grant that $f$ is an homomorphism of groups:

  • for all $g,h \in G$ and $x \in G$ we have $$f(gh)(x)=(gh)x=g(hx)=f(g)(hx)=f(g)(f(h)(x))=f(g)\circ f(h)(x)$$
  • for all $x \in G$ the equality $$f(1)(x)=1x=x$$ holds.

This homomorphism $f$ is the left regular representation (it represents the elements of the group $G$ as symmetries of the same group seen just as a set).

The name left is to distinguish from the (anti)-homomorphism $$f' \colon G \to S(G)$$ that to every $g \in G$ associate $f'(g) \colon G \to G$ such that for every $x \in G$ we have $f'(g)(x) = xg$.

Via the correspondence between actions and homomorphisms in symmetric groups left regular representation are those representation that correspond to left-action of $G$ on itself given by multiplication.

Hope this helps.

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Thank you Giorgio. Just to confirm: So if we have $G = \{1,2,3\}$, then $\pi(G) = S_3$. For example, $S = (1 \ 2 \ 3) \in \pi(G)$ takes $1 \in G$ to 2 - right...? – Tumbleweed Sep 24 '13 at 16:33
@Tumbleweed Well it seems true, but I don't get the relevance of this for the question. – Giorgio Mossa Sep 24 '13 at 16:41

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