Finding irreducible complex representations of $A_{4}$ without character theory We've been working on finding the irreducible representations of finite groups using just the bare definitions of representations, sub-representations, and etc. I'm seeking to use these definitions to classify irreducible representations of $A_{4}$.
This is so far what I've considered, I apologize if this is a bit of a mess
If we start with any representation $V$ of $A_{4}$. We have a normal subgroup $N$ that is generated by $\tau=(12)(34)$ and $\tau'=(13)(24)$ and is isomorphic to the Klein $4$ group. Looking at these generators as invertible linear operators, since $\tau^{2}=\tau'^{2}=1$, we have that the eigenvalues for $\tau$ and $\tau'$ are $\pm 1$, we also have that these operators are simultaneously diagonalizable. We can write the representation $V$ as 
$V=E_{1} \oplus E_{-1} = E'_{1} \oplus E_{1}$
The $E_{i}$ are the eigenspaces for $\tau$ with eigenvalue $i$ and $E'_{i}$ is the same thing for $\tau'$. We wish to construct $A_{4}$ invariant subspaces out of these eigenspaces. For $\sigma=(123)$, we have that $A_{4}=N \sqcup \sigma{N} \sqcup \sigma^{2}{N}$. It then helps to consider the action of $\sigma$ and $\sigma^{2}$ on these eigenspaces. Since $N$ is a normal subgroup we have the following relations
$ \tau \sigma = \sigma \tau' $,
$ \tau \sigma^{2}= \sigma^{2} \tau \tau'$,
$ \tau' \sigma= \sigma \tau \tau'$,
$ \tau' \sigma^{2} = \sigma^{2} \tau$
This tells us that if we have an eigenvector $v$ for $\tau'$, then $\sigma(v)$ will be an eigenvector for $\tau$. This tells us that $\sigma$ will send $E'_{-1} \mapsto E_{-1}$. I am a bit confused on where to go from here and how to find invariant subspaces.
 A: Maybe it would be easier to consider the four 1-dimensional $N$-modules and induce up to $A_4$. That is, let $\epsilon,\epsilon'\in\{+,-\}$, and let $V_{\epsilon,\epsilon'}$ be the 1-dimensional $N$-module with $\tau.v=\epsilon v$ and $\tau'.v=\epsilon' v$ for all $v\in V_{\epsilon,\epsilon'}$. 
Now, consider the $A_4$-module $W_{\epsilon,\epsilon'}$ with basis $\{v,\sigma v,\sigma^2v\}$ where $v$ is a nonzero vector in $V_{\epsilon,\epsilon'}$ (Formally, $W_{\epsilon,\epsilon'}=\mathbb{C}A_4\otimes_{\mathbb{C}N}V_{\epsilon,\epsilon'}$, but never mind). You can now use the defining relations in the post to compute the action of $A_4$ on $W_{\epsilon,\epsilon'}$ (though the 3rd relation should be $\tau'\sigma=\sigma\tau\tau'$). For example, 
\begin{align*}
\sigma.\sigma^iv&=\sigma^{i+1} v,\\
\tau.\sigma v&=(\tau\sigma).v=(\sigma\tau').v=\epsilon'\sigma v\\ 
\tau'.\sigma v&=(\tau'\sigma).v=(\sigma\tau\tau').v=\epsilon\epsilon'\sigma v.
\end{align*}
You can now decompose $W_{\epsilon,\epsilon'}$ into irreducibles. For example, since $N$ acts trivially on $W_{+,+}$, $W_{+,+}$ is a module for $A_4/N\cong C_3$. Hence, $W_{+,+}$ decomposes as the direct sum of three 1-dimensional submodules on which $\sigma$ acts as a third root of unity.
You'll find that the other three modules are irreducible (in fact, they are isomorphic). To see that this gives you everything, you can observe that the left regular module is isomorphic to $W_{++}\oplus W_{+-}\oplus W_{-+}\oplus W_{--}$. As submodules of the left regular module, these summands are cyclically generated by $(1+\tau)(1+\tau')$, $(1+\tau)(1-\tau')$, $(1-\tau)(1+\tau')$ and $(1-\tau)(1-\tau')$, respectively.
