Past exam of the algebra course of the University of Oxford / 2010 / B / 6

A complex character of a representation of a finite group $G$ is the functionwhere $\rho :G\to \mathrm{GL}(V)$ is a finite-dimensional complex group representation. The character is a class function because trace is unchanged by conjugation. Its degree is $\chi (1)=\dim V$.

The row orthogonality relations for a character table assert that distinct complex irreducible characters are orthogonal and each has squared norm one:Using the linearity and conjugate-linearity of the character inner product,Since the multiplicities are nonnegative integers, this is

By the Maschke theorem, the restricted representation is a direct sum of irreducible $H$-modules. The row orthogonality relations for a character table identify the multiplicity of ${\psi }_i$ as $d_i=⟨\chi {\downarrow }_H,{\psi }_i{⟩}_H$. ThereforeUsing the preceding character inner product identity and the irreducibility of $\chi$,This is the character restriction norm bound. When $[G:H]=2$, the nonzero sum of squares of integer multiplicities is either one or two. Hence the restriction is irreducible, or is the sum of two distinct irreducible characters, each with multiplicity one. This proves the requested index-two character restriction dichotomy directly; normality of an index-two subgroup is a general group theory fact.

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.