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I have to show that the group action on $A_4$ on the vertices of a tetrahedron gives a reducible representation $\chi$. Show that it also is the direct sum of an irreducible representation of $A_4$ along with the standard one. Here is what I have:

First, I showed that $A_4/N$ is isomorphic to $Z/3Z$.Here,N is the group with elements $ {1,(12)(34),(13)(24),(14)(23)}$. Here, I used the map $(123) -> w^k$, where $w$ is the root of unity for $k = 0,1,2$. This gives three irreducible representations, call them $\chi_1$,$\chi_2$,$\chi_3$. Moreover, we have another irreducible representation $\chi_4$. This is because there are 4 conjugacy classes to $A_4$. I can fill out the table and everything, but I am stuck on how to proceed on the group action part and coming up with a reducible representation by direct sums. Any direction/hints is greatly appreciated.

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Look at the character $\chi$ of the representation. It takes each element of $A_4$ to the number of its fixed points when acting on the vertices, so $1\mapsto 4$, $(12)(34)\mapsto 0$ and $(123)\mapsto1$ etc. The number of times the trivial character $\epsilon$ appears in $\chi$ is $$\left<\epsilon,\chi\right>=\frac1{12}\sum_{\rho\in A_4} \epsilon(\rho)\overline{\chi(\rho)}=\frac1{12}\sum_{\rho\in A_4} \chi(\rho)=1.$$ If we let $\chi'=\chi-\epsilon$ we find $\left<\chi',\chi'\right>=1$, so that $\chi'$ is irreducible,

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  • $\begingroup$ Thanks, I got it. So $1 -> 4$, $ (12)(34) -> 0$, $(123) ->1$, $(132)->1$ under the action of G on A ={a,b,c,d}, the vertices of said tetrahedron. And, so we have a reducible representation so that it decomposes as a direct sum $(4,1,1,0) = (3,0,0,-1) + (1,1,1,1)$, where the representations I have on the right are the permutation and the trivial one. Thanks! $\endgroup$ – LordVader007 May 5 '18 at 3:21

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