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Suppose first that $N$ is the Frobenius kernel. An element $g\notin N$ fixes a point in the Frobenius action. If $n\in C_N(g)$, then $n$ maps that point to another point fixed by $g$. Since $g$ has only one fixed point and $N$ acts regularly, $n=1$. ThusNo nonidentity conjugacy class of $N$ can be fixed by such a $g$. If the class of $n\ne 1$ were fixed, $gn{g}^{-1}=vn{v}^{-1}$ for some $v\in N$, so $v^{-1}g\notin N$ would centralize $n$, a contradiction. Conjugation by $g$ consequently fixes just the identity class. By Brauer's permutation lemma it fixes just one irreducible character, necessarily $1_N$.

For every nonprincipal $\phi \in \mathrm{Irr}(N)$, its inertia group of a character is therefore $I_G(\phi )=N$. Since $N$ is normal, the restriction of the induced character is the sum of its conjugates, counted over $G/N$:Frobenius reciprocity and character orthogonality giveAn actual character of norm one is irreducible. Every nonprincipal irreducible character of $N$ therefore induces irreducibly to $G$.

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.

For a character $\chi$ of a representation of $H\le K$, the induced character isThe sum is zero when no conjugate of $g$ meets $H$.

Write $g=(a,b)$ for the affine map $z\mapsto az+b$. Conjugation by $(a,0)$ sends translation by $b$ to translation by $ab$, so it permutes transitively the nontrivial characters of $U$. Since $U$ is normal, the induced character is zero outside $U$. On a translation by $b$ it is the sum of all conjugates of $\psi$, henceFor $b\ne 0$ the sum of the nontrivial additive characters is minus one, since the sum including the trivial character is zero. Its character norm isBy the irreducibility criterion for characters, $\rho ={\mathrm{Ind}}_U^G\psi$ is irreducible of dimension $p-1$. All choices of nontrivial $\psi$ give the same representation.

Use the defining property of the Frobenius complement: $H\cap H^x=\{1\}$ whenever $x\notin H$. The induced character formula, extended linearly to any class function, isFor $h\in H\setminus \{1\}$, a summand requires $h\in H\cap xH{x}^{-1}$. The Frobenius intersection property forces $x\in H$. All $|H|$ remaining terms equal $\theta (h)$ since $\theta$ is a class function on $H$, so ${\theta }^G(h)=\theta (h)$. At the identity,Together these giveThe zero identity value is essential: at nonidentity elements the intersection argument already gives the equality, whereas induction multiplies the identity value by the subgroup index.

We use complex representations. By Maschke's theorem, restriction to any finite subgroup is a semisimple module, so it has a canonical isotypic decomposition. In particular, the $\theta$-isotypic component $W_{\theta }$ is the sum of all copies of the irreducible $N$-module with character $\theta$. The central character idempotentprojects onto this component. These projections preserve every $N$-submodule, which consequently decomposes as the direct sum of its intersections with the isotypic components.

Let $W$ afford $\xi \in \mathrm{Irr}(T\mid \theta )$. The component $W_{\theta }$ is nonzero. Since $T=I_G(\theta )$ fixes $\theta$, it preserves $W_{\theta }$. Irreducibility of $W$ as a $T$-module therefore gives $W=W_{\theta }$: its restriction to $N$ consists entirely of copies of $\theta$.

Form the induced representationwhere $\mathcal{R}$ represents the left cosets $G/T$. Because $N◃G$, for $n\in N$ we haveThus $g\otimes W$ is $N$-isotypic of type ${\theta }^g$, where ${\theta }^g(n)=\theta (g^{-1}ng)$. These types are distinct for distinct cosets $gT$, precisely by the definition of the inertia group of a character $T$.

Let $Y$ be a nonzero $G$-submodule of $V$. Its isotypic decomposition as an $N$-module shows that it meets some $g\otimes W$ nontrivially. Acting by $g^{-1}$ gives $Y\cap (1\otimes W)\ne 0$. This intersection is a $T$-submodule of the irreducible $W$, hence equals $1\otimes W$. The $G$-translates of that component span $V$, so $Y=V$. The induced representation is therefore irreducible, and its restriction contains $\theta$. Its induced character satisfiesThis proves the irreducibility assertion directly from isotypic components; no form of the correspondence being proved has been assumed.

The induced character construction extends any class function $\psi$ on $H$ byFrobenius reciprocity for class functions is $⟨{\mathrm{Ind}}_H^G\psi ,\chi {⟩}_G=⟨\psi ,{\mathrm{Res}}_H^G\chi {⟩}_H$. It follows directly by substituting this sum into the left inner product, writing $g=xh{x}^{-1}$, and using the conjugacy invariance of $\chi$. If $\psi$ is a character, its inner product with every restricted irreducible character is a nonnegative integer, a multiplicity in its complete reducible decomposition. Expanding in the orthonormal irreducible-character basis therefore makes $\mathrm{Ind}\psi$ a nonnegative integer sum of irreducible characters: it is a character of $G$.

If $U$ affords $\psi$, a concrete induced representation is $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}U$, with $G$ acting by left multiplication on the first factor. Choosing coset representatives decomposes it into $[G:H]$ copies of $U$. An element $g$ permutes these copies; only fixed cosets contribute trace, and the contribution from $xH$ is $\psi (x^{-1}gx)$. Summing fixed cosets gives exactly the displayed induced class function.

Each $H$-conjugacy class represented by $x_i$ has $|H|/|C_H(x_i)|$ elements. For every element of that class, the number of conjugators from $g$ to it is $|C_G(g)|$. Grouping the induced character formula by these classes therefore gives

In the example, $a$ has order four, $b$ has order two, and $bab=a^{-1}$, so $H\cong D_8$. Its five conjugacy classes are $1$, $a^2$, $\{a,a^3\}$, $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$, of sizes $1,1,2,2,2$. The four one-dimensional irreducible characters are determined by $\psi (a)=\epsilon$ and $\psi (b)=\eta$, each sign independently chosen. The remaining two-dimensional irreducible character comes from the plane action of the square:Their squared degrees sum to eight, so this is the full character table.

In $S_4$, use class representatives $1,(12),(12)(34),(123),(1234)$. Their intersections with $H$ are respectively $\{1\}$, $\{b,a^{2}b\}$, $\{a^2\}\cup \{ab,a^{3}b\}$, the empty set, and $\{a,a^3\}$. The corresponding $S_4$ centralizer orders are $24,4,8,3,4$, while the noncentral $H$ centralizers have order four. The induced values are consequentlyIn the same $S_4$ class order, all five induced irreducible characters are

In the induced character formula above, there are no $x\in G$ with $x^{-1}gx\in H$ when the conjugacy class of $g$ misses $H$. The sum is empty, so $({\mathrm{Ind}}_H^G\psi )(g)=0$.

For a character $\psi$ of a subgroup $H\le G$, the induced character isIt is the character of the induced representation $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}{V}_{\psi }$, and has degree $[G:H]\psi (1)$. The character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_g\alpha (g)\overline{\beta (g)}$. Substituting the displayed formula and putting $g=xh{x}^{-1}$ yieldsThus Frobenius reciprocity isNow assume index two, so $H$ is normal. Choose $g\notin H$ and write ${\psi }^g(h)=\psi (g^{-1}hg)$. Directly from the induction formula,If ${\psi }^g\ne \psi$, this norm is one, so induction is irreducible of degree $2d$: $\psi$ is monogamous. If ${\psi }^g=\psi$, the norm is two. Since a norm is the sum of squared nonnegative integer multiplicities, induction is the sum of two distinct irreducibles, each once. Each is related to $\psi$, so its degree is at least $d$ by restriction and reciprocity; their degrees sum to $2d$, forcing both to have degree $d$. Thus $\psi$ is bigamous.

Every irreducible $\chi$ of $G$ restricts to a nonzero sum of irreducibles. Choose a constituent $\psi$. If $\psi$ is bigamous, $\chi$ is one of its two degree-$d$ partners; since its restriction already contains a degree-$d$ constituent, $\mathrm{Res}\chi =\psi$. If $\psi$ is monogamous, $\chi =\mathrm{Ind}\psi$ and $\mathrm{Res}\chi =\psi +{\psi }^g$, giving precisely two distinct monogamous partners of equal degree.

The complex irreducible character degrees of $A_5$ are $1,3,3,4,5$, with squared sum $60$. Conjugation by the outside coset preserves degrees, so it fixes the unique degree-$1$, degree-$4$ and degree-$5$ characters and either fixes or swaps the two degree-$3$ characters. The two possibilities for $G$ areBoth lists have squared sum $120=|G|$. They occur for $A_5\times C_2$ and $S_5$ respectively; in $S_5$, odd conjugation swaps the two classes of $5$-cycles and hence the two degree-$3$ characters.