Can someone tell me in dummy terms what Left and Right regular representations are. In my book it just says that the left regular representation is the map f in the Cayley's Theorem proof, but I just don't understand what are they? Why we need to find them?
 A: Consider a real vector space $V_G$ whose basis is elements of $G$: i.e., elements in $V_G$ are of the form $\sum_{g \in G} a_g g$, where $a_g \in \Bbb R$. Then $G$ acts on this space by the map $$h\left(\sum_{g \in G} a_g g\right) = \sum_{g \in G} a_g hg.$$ This is the left regular representation. The right one is defined analagously, by $$h\left(\sum_{g \in G} a_g g\right) = \sum_{g \in G} a_g gh^{-1}.$$ (We need to multiply by $h^{-1}$ if we want this to be a group action, or else $(gh)(v) \neq g(hv)$!)
If you define a group representation to be a homomorphism $G \to GL_n(\Bbb R)$ for some $n$, you get one by sending an element $g \in G$ to the vector space transformation it induces on $V_G \cong \Bbb R^{|G|}$; i.e., $g \mapsto (v \mapsto gv)$. The transformations $v \mapsto gv$ are linear and invertible, hence are elements of $GL(V_G)$.
A: Mike's answer is perfectly reasonable but I prefer to look at it by example because it becomes clearer where it comes from. I think the most natural setting is group actions. A group always acts on itself by multiplication, i.e. if $g\in G$, then we can look at what happens as we multiply elements of $G$ by $g$ on the left or right. Take for instance $\Bbb Z_3$ and $1$. Let the symbol $\rhd$ denote action, then
$$1\rhd 0 = 1+0 = 1$$
$$1\rhd 1 = 1+1 = 2$$
$$1\rhd 2 = 1+2 = 0$$
Then we can attribute a matrix to $1$ if we let $0$, $1$ and $2$ be "basis vectors" in some sense. (Here I will let $0 \Longleftrightarrow \left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right)$, $1 \Longleftrightarrow \left(\begin{array}{c}0 \\ 1 \\ 0\end{array}\right)$ and $2 \Longleftrightarrow \left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right)$). The above is just a set of linear equations, which is ultimately the reason we consider such a construction - it looks like a matrix-vector product. Doing so, we get that
$$[1] = \left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right).$$
So that we obtain for instance
$$1\rhd 0 \Longrightarrow \left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right)\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right) = \left(\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right)$$
but we attribute this vector to $1$, which is in exact agreement with what we noticed above. You can repeat this process in general and the exact way by which to do that is Mike's approach.
Underlying this is what is known as the group ring. The group ring is formal linear combinations of group elements with scalars coming from your underlying field. (This is what Mike has done.) If your field is $\Bbb{C}$, the group ring is denoted $\Bbb{C}[G]$. More explicitly, $\Bbb{C}[G] = \{\alpha_1 g_1+\cdots+\alpha_n g_n:\alpha_i\in\Bbb{C},g_j\in G\}$. If this notation is somewhat unsettling for you, you could instead look at it as being an $n$ dimensional vector space with coefficients in $\Bbb{C}$, but the vector space is also equipped with "multiplication" because we can multiply group elements.
As for why we look to the left and right regular representations, it is because (if I'm not mistaken..) every irreducible representation is contained inside the left and right regular representations. So to find out what all of the irreducibles are for a group, we just need to find out what the irreducibles are for the left (or right) regular representation.
A: The regular representation usually just refers to the group (or whatever you are looking at) acting on itself by left or right multiplication.
A: As @Grifulkin states, the left (resp. right) regular action is simply the group acting on itself by left (resp. right) multiplication.  An action of a group $G$ on a set $X$ is equivalent to a homomorphism $G \to \text{Sym}(X)$ where $\text{Sym}(X)$ is the group of all permutations of $X$.  This homomorphism is called a permutation representation.
I'm not sure why the other responses are so focused on vector spaces, since Cayley's Theorem makes no mention of them.  Cayley's Theorem states that a finite group of order $n$ can be embedded as a subgroup of the symmetric group $S_n$.  In this post, I illustrate this fact by showing that the elements of the Klein 4-group can be represented by permutations in $S_4$:
First, we choose a numbering of the elements of $V$.  Somewhat arbitrarily, I choose to label $1,a,b,ab$ by $1,2,3,4$, respectively.  First note that $1$ must act as the identity permutation.  We observe that $a$ acts by left multiplication by sending
\begin{align*}
a: 1 &\mapsto a\\
a &\mapsto a^2 = 1\\
b &\mapsto ab\\
ab &\mapsto b \, .
\end{align*}
Recalling our numbering we see that the permutation $\sigma_a$ corresponding to $a$ sends
\begin{align*}
\sigma_a: 1 &\mapsto 2\\
2 &\mapsto 1\\
3 &\mapsto 4\\
4 &\mapsto 3
\end{align*}
so $\sigma_a = (1\ 2)(3\ 4)$.  Proceeding similarly with $b$ and $ab$, we find $\sigma_b = (1\ 3)(2\ 4)$ and $\sigma_{ab} = (1\ 4)(2\ 3)$.
Note that choosing a different numbering for the elements of $V$ will yield different permutations.  In fact changing the numbering conjugates all the permutations.
A: The left and right regular representations of a group $G$ are the isomorphic images of $G$ into $\operatorname {Sym}(G)$, via respectively the embeddings $a\mapsto (g\mapsto ag)$ and $a\mapsto (g\mapsto ga^{-1})$. If we name them $\Lambda $ and $\text{R}$, respectively, then it turns out that the center of $G$, $Z(G)$, is isomorphic to $\Lambda \cap\text{R}$. Moreover, $\Lambda \text{R}=\text{R}\Lambda $, and hence $\Lambda \text{R}\le\operatorname {Sym}(G)$. Finally, $\operatorname {Inn}(G)=\operatorname {Aut}(G)\cap\Lambda \text{R}$, thus giving to the "inner" adjective a kind of plausibile base (considering $\Lambda \text{R}$ as "the great image of $G$" in $\operatorname {Sym}(G)$). So, left and right regular representations are not just abstract entities, but they have set-wise connections with group-theoretic building blocks such as, at least, $Z(G)$ and $\operatorname {Inn}(G)$.
