Solution
= Solution
Write $o(\theta)$ for the order of the <determinant character> of an <irreducible character> $\theta$. An extension has the same degree and restricts exactly to $\theta$, rather than just containing it as a constituent.
\b[Extension theorem.] <Thompson's coprime character extension theorem> says that if $N\triangleleft G$, $\theta\in\operatorname{Irr}(N)$ is $G$-invariant, and $\gcd(\theta(1),[G:N])=1$, then $\theta$ extends to $G$ exactly when its <determinant character> extends to a <linear character> of $G$. In particular, if also $\gcd(o(\theta),[G:N])=1$, there is a unique extension with determinant order coprime to the index. For a <p-group> quotient, the version used below gives a canonical extension whenever both the degree and the determinant order are prime to $p$. This theorem is an assertion of extendibility, not the false statement that every invariant <irreducible character> extends.
\b[The p-nilpotency lemma.] Define
$$
s_p(G)=\sum_{\chi\in\operatorname{Irr}(G),\ p\nmid\chi(1)}\chi(1)^2,\qquad
h_p(G)=|G:G'|_p,
$$
where $m_p$ is the largest power of $p$ dividing an integer $m$. <Isaacs' character-degree criterion for p-nilpotency> asserts
$$
\boxed{h_p(G)\mid s_p(G),\qquad
G\text{ has a normal }p\text{-complement}\ \Longleftrightarrow\ p\nmid\frac{s_p(G)}{h_p(G)}}.
$$
Equivalently, $s_p(G)_p=h_p(G)$. We prove both the divisibility and the criterion.
Let $N=O^p(G)$ be the <p-residual of a finite group>, the least <normal subgroup> with <p-group> quotient. It can also be described as the <subgroup> generated by all p-prime-order elements: every homomorphism to a <p-group> kills those elements, and after quotienting by their generated <subgroup> the image of every element has p-power order, so the quotient is a <p-group> by Cauchy's theorem.
The residual is p-perfect, meaning $O^p(N)=N$. Indeed $O^p(N)$ is characteristic in $N$, hence normal in $G$, and $G/O^p(N)$ is an extension of the two <p-groups> $N/O^p(N)$ and $G/N$. Minimality of $N$ forces equality. Consequently $N/N'$ has order prime to $p$, since otherwise it would have a quotient of order $p$. Every <linear character> of $N$, in particular every <determinant character> of $N$, therefore has order prime to $p$.
Let $\mathcal T$ consist of the $G$-invariant <irreducible characters> $\theta$ of $N$ of degree prime to $p$. Each extends to an <irreducible character> $\widehat\theta$ of $G$ by the extension theorem. The <irreducible characters> of $G$ lying over $\theta$ are exactly
$$
\widehat\theta\,\nu,\qquad \nu\in\operatorname{Irr}(G/N).
$$
Here is the <representation> argument for this description. If a <representation> lies over the invariant $\theta$, its $N$-module is copies of the module for $\theta$. Identify it with that module tensored with its multiplicity space. Dividing the $G$-action by the fixed extending action on the first factor leaves an action on the multiplicity space, trivial on $N$. Irreducibility is equivalent to irreducibility of that quotient action, proving the displayed parametrization.
Every irreducible degree of the <p-group> $G/N$ is a power of $p$, since it divides the group order. Thus the p-prime-degree <characters> lying over $\theta$ are exactly those with $\nu$ linear, and there are
$$
|(G/N):(G/N)'|=|G:NG'|=h_p(G)
$$
of them. For the last equality, $G/NG'$ is the largest abelian <p-group> quotient of $G$, hence the p-primary part of its <abelianization>. In particular all these <characters> have degree $\theta(1)$.
Conversely, for any $\chi\in\operatorname{Irr}(G)$ of p-prime degree, Clifford theory writes its restriction to $N$ as $e$ times the sum of a single orbit of <irreducible characters>. The orbit size divides $[G:N]$ and also divides $\chi(1)$, so it is one; the constituent $\theta$ is invariant and has p-prime degree. Hence $\theta\in\mathcal T$. The preceding extension parametrization now forces the quotient <character> to be linear. We have therefore obtained the exact identity
$$
\boxed{s_p(G)=h_p(G)\sum_{\theta\in\mathcal T}\theta(1)^2}.
$$
This proves the required divisibility without assuming p-nilpotency.
The <p-group> $G/N$ acts on $\operatorname{Irr}(N)$ by conjugation. Every nonfixed orbit has size divisible by $p$, and degrees are constant on each orbit. Fixed <characters> whose degrees are divisible by $p$ also contribute zero modulo $p$. The <sum of squares of irreducible degrees> consequently gives
$$
\frac{s_p(G)}{h_p(G)}=\sum_{\theta\in\mathcal T}\theta(1)^2
\equiv\sum_{\theta\in\operatorname{Irr}(N)}\theta(1)^2
=|N|\pmod p.
$$
The quotient is prime to $p$ exactly when $N$ has order prime to $p$. Since $G/N$ is a <p-group>, that is exactly the assertion that $N$ is a <normal p-complement>. Conversely, any normal p-complement equals $O^p(G)$ because it contains all p-prime-order elements and has <p-group> quotient. This finishes the lemma.
\b[Detection from the <group algebra>.] For <finite groups> $G$ and $H$ with isomorphic complex <group algebras>, the <Artin–Wedderburn theorem> gives
$$
\mathbb C[G]\cong\bigoplus_{\chi\in\operatorname{Irr}(G)}M_{\chi(1)}(\mathbb C).
$$
The simple matrix-block sizes and their multiplicities are determined by the algebra isomorphism. Hence $G$ and $H$ have identical irreducible degree multisets. They have the same order, from the degree-square sum, and the same number of <linear characters>, which equals $|G:G'|$. Thus $s_p$ and $h_p$ coincide for every prime $p$. The lemma shows that one group is p-nilpotent exactly when the other is.
A <finite group> is <nilpotent> exactly when it has a normal p-complement for every prime dividing its order. For completeness, if all those complements exist, set $L_q=\bigcap_{p\ne q}K_p$, where $K_p$ is the normal p-complement. The diagonal map $G/L_q\to\prod_{p\ne q}G/K_p$ is injective, so $[G:L_q]$ is prime to $q$. At the same time, containment in each $K_p$ removes every prime other than $q$ from $|L_q|$. Thus $L_q$ is a normal Sylow q-subgroup. Distinct normal <Sylow subgroups> commute, giving the direct-product decomposition. Conversely, the direct-product decomposition of a finite nilpotent group provides all the complements. We conclude
$$
\boxed{\mathbb C[G]\cong\mathbb C[H]\quad\Longrightarrow\quad
(G\text{ nilpotent}\Longleftrightarrow H\text{ nilpotent})}.
$$
This is the assertion that the <complex group algebra detects p-nilpotency> and hence nilpotency. It needs the multiplicities of the degrees, not just their distinct values.
\b[Normal-complement consequence.] If $p$ divides every nonlinear irreducible degree, the p-prime-degree <characters> are exactly the <linear characters>. Therefore
$$
s_p(G)=|G:G'|,\qquad s_p(G)_p=h_p(G).
$$
The lemma yields a normal p-complement immediately:
$$
\boxed{p\mid\chi(1)\text{ for every nonlinear }\chi\quad\Longrightarrow\quad
G\text{ has a normal }p\text{-complement}}.
$$
This proves <Thompson's character-degree normal complement theorem>, including the abelian case where the hypothesis on nonlinear <characters> is vacuous.