How does one decompose the regular representation of the group How does one decompose the regular representation of  the group $<a>_2*<b>_3$
 in the direct sum of  1-dimensional representations 
I know what is regular representation of  the group (G acts on this basis as follows:
$λ(g)(e_h)=e_{gh}$)
$<a>_2*<b>_3={e, b,b^2, a, ab,ab^2}$
Could any of you help me, please? Sorry  for my poor English
 A: The regular representation of a finite group is the group algebra $K[G]$. As a representation of $G$ it decomposes as $\bigoplus V^{\dim V}$, where $V$ ranges over all irreducible representations of $G$. In particular if $G$ is abelian then all irreducible representations are one-dimensional, and they correspond to the elements of the dual group $\widehat{G}$, i.e. the characters $G\to \Bbb C^\times$ ($\cong {\rm GL}(\Bbb C^1)$) - this can be proven using Schur's lemma (for each $g\in G$, $\rho(g)$ commutes with $\rho(h)$ for all $h\in G$, hence $\rho(g)$ is scalar..)
A nice method of decomposing an algebra $A$ is Artin-Wedderburn: find the central orthogonal idempotents ($\ne0$). Suppose one has a maximal such set $\{e_1,\cdots,e_m\}$. Notice $s=e_1+\cdots+e_m$ is also a central idempotent, and then so is $1-s$; if $s\ne1$ then $\{e_1,\cdots,e_m,1-s\}$ is also a collection of central orthogonal idempotents, but is even bigger contra minimality of $m$, so $s=1$. Therefore
$$A=A1=A(e_1+\cdots+e_m)\subseteq Ae_1+Ae_2\cdots+Ae_m\subseteq A\quad\Rightarrow\quad A=\sum_{i=1}^m Ae_i.$$
Furthermore, the sum is direct; if an element of $A$ is representable in two distinct ways, then by subtracting we find $0$ is a nontrivial linear combination of the $e_i$'s, say with $c_ie_i\ne0$ for some index $i\in\{1,\cdots,m\}$, in which case $0=e_i0=e_i(\sum_{j=1}^m c_je_j)=c_ie_i^2=c_ie_i\ne0$, a contradiction. On top of this, the algebra $A$ cannot be decomposed any further (exercise; use maximality of $m$ plus the fact that direct sums of $k$ rings have $k$ identifiable nonzero central orthogonal idempotents).
Suppose $G$ is abelian. Everything in $\Bbb C[G]$ is central then. We seek $n=|G|$ orthogonal idempotents, so first let's figure out the idempotent part. This condition reads
$$\sum_{g\in G}a_gg=\left(\sum_{g\in G}a_gg\right)^2=\sum_{g\in G}\left(\sum_{h\in G}a_ha_{gh^{-1}}\right)g.$$
Identification of coefficients harks back to characters; they satisfy the relations
$$\sum_{h\in G}\chi(h)\chi(gh^{-1})=|G|\chi(g)$$
(since $\chi(h)\chi(gh^{-1})=\chi(h)\chi(g)\chi(h)^{-1}=\chi(g)$) which is almost what we want except for the unwanted factor of $|G|$ present; normalizing yields the idempotents
$$e=\frac{1}{|G|}\sum_{g\in G}\chi(g)g$$
as $\chi$ varies over characters $\in\widehat{G}$. Are they orthogonal? Yes:
$$e_\chi e_\psi=\frac{1}{|G|^2}\sum_{g,h\in G}\chi(g)\psi(h)gh=\frac{1}{|G|^2}\sum_{g\in G}\left(\sum_{h\in G}\chi(h)\psi(gh^{-1})\right)g=0$$
since $\sum_h\chi(h)\psi(gh^{-1})=\psi(g)\sum_h (\chi\cdot\psi^{-1})(h)=0$ if $\chi\ne\psi$ by character theory. Therefore,
$$\Bbb C[G]\cong \bigoplus_{\chi\in\widehat{G}}\left(\sum_{g\in G}\chi(g)g\right)\Bbb C $$

In the case of $G=C_2\times C_3$, the characters $\chi\in\widehat{G}$ are easy to describe. As $\widehat{G}\cong\widehat{C_2}\times\widehat{C_3}$, every character is a product of a homomorphism $C_2\to\Bbb C^\times$ and a homomorphism $C_3\to\Bbb C^\times$, which simply send generators to the appropriate roots of unity.
