 Solution

Use the column order $1,(123),(12)(34),(12345),(13452)$ in $A_5$. The two restrictions of degree four coincide and equal ${\theta }_4=(4,1,0,-1,-1)$; those of degree five coincide and equal ${\theta }_5=(5,-1,1,0,0)$. The centralizer orders giveThus both are irreducible characters. The remaining given restriction is $\eta =(6,0,-2,1,1)$, andIt splits as two distinct irreducible characters. Neither is $1$, ${\theta }_4$ or ${\theta }_5$, since their character inner products with $\eta$ are zero. A finite group has as many complex irreducible characters as conjugacy classes, so these five exhaust them. The sum of squares of irreducible degrees is $|A_5|=60$. The first three degrees contribute $1+16+25=42$, leaving $18$ for the two constituents, whose degrees sum to six. Hence both have degree three.

Write one constituent as $\theta =(3,x,y,u,v)$. All five conjugacy classes are closed under inversion, so all its values are real: $\chi (g^{-1})=\overline{\chi (g)}$. Orthogonality with $1$, ${\theta }_4$ and ${\theta }_5$ givesSolving gives $x=0$, $y=-1$ and $u+v=1$. The squared norm one gives $9/60+1/4+(u^2+v^2)/5=1$, so $u^2+v^2=3$ and $uv=-1$. Thus $u,v$ are the roots of $t^2-t-1$, namely $\phi =(1+\sqrt{5})/2$ and $\bar{\phi }=(1-\sqrt{5})/2$. Here the bar labels the other real root, rather than complex conjugation. The other constituent is $\eta -\theta$, so the two last values are interchanged.

The Character table of the alternating group A5 is thereforeThe conjugacy class sizes are $1,20,15,12,12$. A normal subgroup is a union of entire conjugacy classes containing the identity, and its order divides $60$ by Lagrange's theorem. A nontrivial such union has size at least $13$; the only proper divisors of $60$ at least $13$ are $15,20,30$. Subtracting the identity would require a sum of some of $20,15,12,12$ equal to $14,19,29$, respectively, none of which is possible. Thus $A_5$ is simple.