Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 2 / 5

Let $V$ afford the given group representation of $H$. Its induced representation is the left $G$-moduleChoose representatives $t$ of the left cosets $G/H$; then it is the direct sum of the spaces $t\otimes V$. If $gt=t^{\prime }h$ with $h\in H$, the action sends $t\otimes v$ to $t^{\prime }\otimes \Psi (h)v$. This is independent of the representatives up to equivalence, and its dimension is $[G:H]\dim V$.

Only cosets fixed by $g$ contribute to its trace. Their contribution is $\psi (t^{-1}gt)$, giving the induced character formulaFor an arbitrary class function $f$ on $H$, the same averaging formula defines its induced class function on $G$. Equivalently, extend $f$ by zero outside $H$ and average $|H{|}^{-1}{\sum }_{x}f(x^{-1}gx)$. This definition is linear and does not require $f$ to be an actual character.

For a complex character $\psi$ of a finite group $G$ afforded by a representation $\rho$, its determinant character isMultiplicativity of the determinant makes this a linear character; it does not mean the determinant of the scalar value $\psi (g)$. The definition depends only on the character because characters determine the equivalence class of a finite-dimensional complex representation. In particular the determinant character is trivial on the derived subgroup $G^{\prime }$.

Frobenius reciprocity is the adjointness of character induction and restriction: for class functions $f$ on $H$ and $u$ on $G$,For characters this says that the multiplicity of an irreducible $G$-character in an induced character equals the corresponding multiplicity after restriction. To prove the identity directly, insert the induced class function formula and interchange the two finite sums. For each $x$, substitute $g=xh{x}^{-1}$ with $h\in H$. Since $u$ is constant on conjugacy classes,This proves the theorem for all class functions, not just irreducible characters.

For the group-theoretic conclusion, suppose that $p$ divides $|G^{\prime }\cap Z(G)|$. A central element of order $p$ lies in every Sylow p-subgroup: adjoining it to a Sylow subgroup gives a p-subgroup, so maximality forces it into that subgroup. Hence there is a subgroup $Q\le P\cap G^{\prime }\cap Z(G)$ of order $p$. Choose a nontrivial linear character $\lambda$ of $Q$.

Since $P$ is abelian, $\lambda$ extends to a linear character $\mu$ of $P$. One direct justification is to extend from an abelian subgroup $A$ to $⟨A,x⟩$ one generator at a time: if $d$ is the least positive exponent with $x^d\in A$, choose a complex $d$th root of $\lambda (x^d)$ as the value at $x$. The formula on $a{x}^j$ is then well defined and multiplicative. Iteration yields the claimed extension to $P$.

The induced character ${\mu }^G$ has degree $[G:P]$, which is prime to $p$. In its decomposition into irreducible characters, at least one constituent $\chi$ must have degree prime to $p$; otherwise the total degree would be divisible by $p$. Frobenius reciprocity also gives $⟨{\chi }_P,\mu {⟩}_P>0$ for this constituent. Restriction to the central subgroup $Q$ forcesIndeed, the induced representation itself restricts to $Q$ as copies of $\lambda$, because conjugation by every coset representative is trivial on $Q$; alternatively this follows from Schur's lemma and the positive multiplicity after restriction.

Each $q\in Q$ consequently acts in the representation affording $\chi$ as the scalar $\lambda (q)$. Taking its determinant character givesThe left side is trivial since $Q\le G^{\prime }$ and every linear character kills $G^{\prime }$. The right side is nontrivial since $\lambda$ has order $p$ and $p\nmid \chi (1)$. This contradiction provesThe argument is the central commutator has no torsion from an abelian Sylow subgroup determinant obstruction.