Let $G=A_4$ be the alternating group on 4 letters, and let $R = \mathbb{C}[G]$. Then
$$\mathbb{C}[G] = U\oplus U' \oplus U'' \oplus V^{\oplus 3},$$
where $U,U',U''$ are the three 1-dimensional irreducible representations and $V$ the one arising from the standard representation of $S_4$.
We want to:
- find idempotents $\epsilon$ for each of $U,U',U''$ that realize these representations as left ideals of $\mathbb{C}[G]$.
- find a subrepresentation $W\subset R$ which is isomorphic to $V$, and express $W = \mathbb{C}[G]\epsilon\subset R,$ where $\epsilon$ is an idempotent.
I know how to use Young diagrams to find idempotents for each irreducible representation of $S_4$. But how can we use the Young diagrams to find these things for $A_4$?
For the first part, note that $A_4$ has a unique nontrivial proper normal subgroup $$K_4:=\big\{(),(1\;2)(3\;4),(1\;3)(2\;4),(1\;4)(2\;3)\big\}\,.$$ There is a group isomorphism $\varphi:(A_4/K_4)\to C_3$, where $C_k$ is the cyclic group of order $k$. A representation $\rho:C_3\to \text{GL}(X)$ of $C_3$ can be made into a representation $\tilde{\rho}: A_4\to \text{GL}(X)$ of $A_4$ by setting $$\tilde{\rho}:=\rho\circ\varphi\circ\kappa\,,$$ where $\kappa:A_4\to (A_4/K_4)$ is the canonical projection. Note that $C_3$ has three nonisomorphic irreducible representations, all of which are $1$-dimensional. The irreducible representations of $C_3=\langle g\rangle$ are $\chi_j:C_3\to\text{GL}(C)$ sending $g\mapsto \exp\left(\frac{2\pi\text{i}j}{3}\right)$ for $j=0,1,2$. We may identify $C_3$ with the subgroup $\big\langle (1\;2\;3)\big\rangle$ of $A_4$, and set $g:=(1\;2\;3)$.
We now know what to look for. The three nonisomorphic $1$-dimensional representations of $A_4$ must be given by $\tilde{\chi}_j$ for $j=0,1,2$. First, the left ideal $U$ which is given by the representation $\tilde{\chi}_0$ of $R$ is clearly generated by $\sum\limits_{\sigma\in A_4}\,\sigma$. We must rescale this element to get an idempotent $\epsilon_U\in R$, which is $$\epsilon_U:=\frac{1}{|A_4|}\,\sum\limits_{\sigma\in A_4}\,\sigma=\frac{1}{12}\,\sum\limits_{\sigma\in A_4}\,\sigma\,.$$
Let $U'$ and $U''$ be the $1$-dimensional left ideal of $R$ corresponding to the representations $\tilde{\chi}_1$ and $\tilde{\chi}_2$, respectively, with the corresponding idempotent elements $\epsilon_{U'}$ and $\epsilon_{U''}$. Now, note that $K_4$ must act trivially on $U'$ and $U''$. Ergo, we have $$\epsilon_{U'}=a_1\,s+b_1\,(1\;2\;3)\,s+c_1\,(1\;3\;2)\,s$$ and $$\epsilon_{U''}=a_2\,s+b_2\,(1\;2\;3)\,s+c_2\,(1\;3\;2)\,s$$ for some $a_1,a_2,b_1,b_2,c_1,c_2\in\mathbb{C}$. Here, $$s:=\sum_{\sigma\in K_4}\,\sigma\,.$$ We clearly have $a_j=\exp\left(\frac{2\pi\text{i}j}{3}\right)\,b_j$ and $b_j=\exp\left(\frac{2\pi\text{i}j}{3}\right)\,c_j$ since $g=(1\;2\;3)$ acts on $U'$ and $U''$ via multiplications by $\exp\left(\frac{2\pi\text{i}}{3}\right)$ and $\exp\left(\frac{4\pi\text{i}}{3}\right)$, respectively. That is, $$\epsilon_{U'}=c_1\,\Biggl(\exp\left(\frac{4\pi\text{i}}{3}\right)\,s+\exp\left(\frac{2\pi\text{i}}{3}\right)\,(1\;2\;3)\,s+(1\;3\;2)\,s\Biggr)$$ and $$\epsilon_{U''}=c_2\,\Biggl(\exp\left(\frac{2\pi\text{i}}{3}\right)\,s+\exp\left(\frac{4\pi\text{i}}{3}\right)\,(1\;2\;3)\,s+(1\;3\;2)\,s\Biggr)\,.$$ Because $\epsilon_{U'}$ and $\epsilon_{U''}$ are idempotent, $$c_j=\frac{1}{|K_4|\,\Biggl(3\,\exp\left(\frac{3\pi\text{i}j}{3}\right)\Biggr)}=\frac{\exp\left(\frac{2\pi\text{i}j}{3}\right)}{12}\text{ for }j\in\{1,2\}\,.$$ Ergo, $$\epsilon_{U'}=\frac{1}{12}\,\Biggl(1+\exp\left(\frac{4\pi\text{i}}{3}\right)\,(1\;2\;3)+\exp\left(\frac{2\pi\text{i}}{3}\right)\,(1\;3\;2)\Biggr)\,\sum_{\sigma\in K_4}\,\sigma$$ and $$\epsilon_{U''}=\frac{1}{12}\,\Biggl(1+\exp\left(\frac{2\pi\text{i}}{3}\right)\,(1\;2\;3)+\exp\left(\frac{4\pi\text{i}}{3}\right)\,(1\;3\;2)\Biggr)\,\sum_{\sigma\in K_4}\,\sigma\,.$$ (Observe that $U$, $U'$, and $U''$ are in fact two-sided ideals of $R$ isomorphic to $\mathbb{C}$.)
For the second part, we can take $W$ to be generated by $$p:=1+(1\;2)(3\;4)-(1\;3)(2\;4)-(1\;4)(2\;3)=\big(1+(1\;2)(3\;4)\big)\,\big(1-(1\;3)(2\;4)\big)\,.$$ Let $q:=(1\;2\;3)\,p$ and $r:=(1\;3\;2)\,p$. Note that $K_4$ acts on $p$ by scalar multiple (the scalars involved are $\pm1$). Since $A_4=C_3K_4$ and $K_4$ is normal in $A_4$, $W$ is indeed spanned by $p$, $q$, and $r$, whence it is a $3$-dimensional left ideal of $R$. As $p^2=4p$, we may take the idempotent $\epsilon_W$ to be $$\epsilon_W:=\frac{1}{4}p=\frac{1}{4}\,\big(1+(1\;2)(3\;4)\big)\,\big(1-(1\;3)(2\;4)\big)$$ so that $W=R\epsilon_W$.
We can also define $W'$ and $W''$ to be the left ideals generated by $p'$ and $p''$, respectively, where $$p':=1+(1\;3)(2\;4)-(1\;2)(3\;4)-(1\;4)(2\;3)=\big(1+(1\;3)(2\;4)\big)\,\big(1-(1\;2)(3\;4)\big)$$ and $$p'':=1+(1\;4)(2\;3)-(1\;2)(3\;4)-(1\;3)(2\;4)=\big(1+(1\;4)(2\;3)\big)\,\big(1-(1\;2)(3\;4)\big)\,.$$ Then, $W'$ is spanned by $p'$, $q':=(1\;2\;3)\,p'$, and $r':=(1\;3\;2)\,p'$, whereas $W''$ is spanned by $p''$, $q'':=(1\;2\;3)\,p''$, and $r'':=(1\;3\;2)\,p''$. The left ideals $W'$ and $W''$ are generated by the idempotents $$\epsilon_{W'}:=\frac{1}{4}\,p'=\frac{1}{4}\,\big(1+(1\;3)(2\;4)\big)\,\big(1-(1\;2)(3\;4)\big)$$ and $$\epsilon_{W''}:=\frac{1}{4}\,p''=\frac{1}{4}\,\big(1+(1\;4)(2\;3)\big)\,\big(1-(1\;2)(3\;4)\big)\,,$$ respectively. We can also see that $W$, $W'$, and $W''$ are isomorphic irreducible $3$-dimensional representations of $A_4$. While each of $W$, $W'$, and $W''$ is a left ideal of $R$, the sum $W\oplus W'\oplus W''$ is a simple two-sided ideal of $R$ isomorphic to $\text{Mat}_{3\times 3}(\mathbb{C})$.