Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 19I (Representation Theory)

The row orthogonality relations for a character table state that for irreducible characters $\chi ,\psi$,Equivalently, if $c$ runs through representatives of the conjugacy classes $C$,

Fix an irreducible $\chi$. The sum of the elements in a conjugacy class $C$ is central in $\mathbb{C}G$, so Schur lemma says that it acts in the representation affording $\chi$ by the scalarThis central character value of a conjugacy-class sum is an algebraic integer: the class sum acts by a matrix with integer entries on the regular representation, and ${\omega }_{\chi }(C)$ is one of its eigenvalues. Also $\overline{\chi (c)}$ is an algebraic integer because character values are sums of roots of unity.

Row orthogonality with $\psi =\chi$ now givesThe right-hand side is an algebraic integer. The left-hand side is rational, and every rational algebraic integer is an integer. Therefore

Let $|G|=p^2$. By irreducible character degree divides the group order, each irreducible degree divides $p^2$. A degree is at most $\sqrt{|G|}=p$ by the sum of squares of irreducible degrees. If a degree $p$ occurred, its square would already consume the entire sum $p^2$, leaving no room for the trivial character. Hence every irreducible degree is one. A finite group has only linear irreducible characters exactly when it is abelian, so every group of order $p^2$ is abelian.

For $p=2$, every group of order four is abelian. Now let $p$ be odd. Degree divisibility and the sum-of-squares formula show that every nonlinear irreducible character has degree two: the other divisors $p$ and $2p$ have squares larger than $2p$.

Let $a$ be the number of linear characters and $b$ the number of degree-two characters. The linear characters are precisely the characters of the abelianization, soThe degree-sum formula givesso $a\equiv 2(\mathrm{mod}4)$. Among the divisors of $2p$, this forces $a=2$ or $a=2p$. In the latter case $b=0$ and all degrees are one. In the former,Thus the degrees areunless they are all one.

Suppose a finite simple group had an irreducible representation $\rho$ of degree two. Its kernel is normal and cannot be the whole group, so $\rho$ is faithful. A nonabelian simple group has no nontrivial linear character; consequently $\det \rho$ is trivial and $\rho (G)\subseteq {\mathrm{SL}}_2(\mathbb{C})$.

Since $2\mid |G|$ by degree divisibility, Cauchy's theorem for finite groups gives an involution $x\in G$. A nonidentity matrix of order two in ${\mathrm{SL}}_2(\mathbb{C})$ has both eigenvalues $-1$, hence equals $-I$. Thus $\rho (x)=-I$ is central in $\rho (G)$. Faithfulness makes $x$ central in $G$, contradicting the trivial centre of a nonabelian simple group. Therefore no finite simple group has an irreducible character of degree two.

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.