Finite Averaging and Character-Theory Prerequisites

Put $s=n\cdot 1_k$. For every $t\in k$, distributivity gives $0t=(0+0)t=0t+0t$, so cancellation gives $0t=0$. Since $0\mathrm{\ne }1$, the scalar zero is not invertible. If $s$ is invertible, it is therefore nonzero, and the equivalence in F2 gives $\mathrm{char}(k)\nmid n$.

Conversely, if $\mathrm{char}(k)\nmid n$, F2 gives $s\mathrm{\ne }0$, and the field inverse axiom gives $s^{-1}\in k$ with $s{s}^{-1}=s^{-1}s=1$.

By integer divisibility, $0\mid n$ would mean $n=0q=0$ for some integer $q$, impossible for $n>0$. Thus characteristic zero is included. At $n=1$, the scalar is $1_k$ with inverse $1_k$. Together with the two implications this proves the claim.

Apply F1 to the empty independent subset of $U$ to obtain a finite basis $u_1,\dots ,u_r$. This is independent in $V$, so apply F1 with subspace $V$ to extend it to $u_1,\dots ,u_r,w_1,\dots ,w_s$. Every vector has an expansion in this basis; two expansions agree coefficientwise because their difference is a zero linear combination of an independent family.

Define $P({\sum }_{j=1}^r{a}_j{u}_j+{\sum }_{\mathrm{\ell }=1}^s{b}_{\mathrm{\ell }}{w}_{\mathrm{\ell }})={\sum }_{j=1}^r{a}_j{u}_j$. The uniqueness just proved makes $P$ a well-defined function $V\to V$. If $v,v^{\mathrm{\prime }}$ have $u$-coordinates $a_j,a_j^{\mathrm{\prime }}$, then $cv+d{v}^{\mathrm{\prime }}$ has $u$-coordinates $c{a}_j+d{a}_j^{\mathrm{\prime }}$. Consequently $P(cv+d{v}^{\mathrm{\prime }})={\sum }_j(c{a}_j+d{a}_j^{\mathrm{\prime }})u_j=cP(v)+dP(v^{\mathrm{\prime }})$, so $P$ is linear.

Each $P(v)$ belongs to $U$. For $u\in U$, its basis expression uses only the $u_j$, so $P(u)=u$. Thus every $u\in U$ is in the image, and $\mathrm{im}P=U$. Since $P(v)\in U$, also $P(P(v))=P(v)$.

If $U=0$, then $r=0$ and the formula gives $P=0$. If $U=V$, no added vectors are needed and $P={\mathrm{id}}_V$. If $V=0$, both lists are empty and these formulas coincide. Only two applications of the choice-free finite extension result and finite enumerations were used; no simultaneous choice over an infinite family occurs.

By F1, both $\mathrm{im}P$ and $\ker P$ are subspaces of the finite-dimensional space $V$. Apply F3 to each empty independent subset to obtain finite bases, and concatenate them in the displayed direct-sum order, putting the $r={\dim }_k\mathrm{im}P$ image basis vectors first. For $u=P(v)$, $P(u)=P^2(v)=P(v)=u$; for $w\in \ker P$, $P(w)=0$. Thus the matrix is $\mathrm{diag}(I_r,0)$.

F2 permits this basis for computing trace. Summing the diagonal gives $\mathrm{tr}(P)=\underset{r\text{ terms}}{\underbrace{1_k+\cdots +1_k}}=r\cdot 1_k$. For $r=0$ this is the empty sum $0$; when $V=0$ it also agrees with F2. For $P={\mathrm{id}}_V$, it gives $({\dim }_{k}V)\cdot 1_k$.

If $\mathrm{char}(k)=p>0$, take $V=k^p$. Its identity has rank $p$ and trace $p\cdot 1_k=0$, whereas its zero map has rank $0$ and trace $0$. Therefore equal traces need not imply equal integer ranks, even among idempotents on the same space.

Write $W$ for the invariant functions. The zero function is invariant. If $f,h\in W$ and $a,b\in k$, then $(af+bh)(g\cdot x)=af(g\cdot x)+bh(g\cdot x)=af(x)+bh(x)=(af+bh)(x)$, so $af+bh\in W$. For each $x$, associativity and commutativity of function addition, the zero and negative identities, $a(f+h)=af+ah$, $(a+b)f=af+bf$, $(ab)f=a(bf)$ and $1_{k}f=f$ are the corresponding field equalities evaluated at $x$. Equality at every $x$ is equality of functions. This verifies all the axioms in F3.

By F2, $x$ and $g\cdot x$ belong to the same orbit. Consequently ${\mathbf{1}}_O(g\cdot x)={\mathbf{1}}_O(x)$, so every orbit indicator lies in $W$. If $f\in W$, its values at any two points of one orbit coincide by F2 and invariance. Each orbit is nonempty, so there is a unique scalar $c_O$ such that $f(x)=c_O$ for all $x\in O$. This defines $c_O$ by its unique value and does not choose orbit representatives.

Define $h={\sum }_{O\in \mathcal{O}}{c}_O{\mathbf{1}}_O$ using F4 in the additive vector space $W$. Evaluation of a finite sum is the sum of its values, by the recursive pointwise addition. At a point $x$, exactly one indicator is $1_k$, namely that of its orbit, and every other term vanishes. Thus $h(x)=c_{G\cdot x}=f(x)$, so $h=f$ and the indicators span $W$.

Suppose ${\sum }_{O\in \mathcal{O}}{a}_O{\mathbf{1}}_O=0$. Fix any orbit $O$ and one $x\in O$, possible because $O$ is nonempty. Evaluation at $x$ gives $a_O=0$. This holds for each orbit separately, so no simultaneous choice is required. Also different orbits have different indicators by evaluation on either orbit and $1_k\mathrm{\ne }{0}_k$. Hence these $\mathrm{\mid }\mathcal{O}\mathrm{\mid }$ vectors are independent and, by F5 and step 2.1, form a basis, giving the asserted dimension.

On $X=G$, set $g\cdot x=gx{g}^{-1}$. Then $ex{e}^{-1}=x$ and $(gh)x(gh{)}^{-1}=g(hx{h}^{-1})g^{-1}$, so this is an action by F1. F6 identifies its orbits as the conjugacy classes. Invariance is precisely constancy on conjugacy classes, the definition of a class function here. If $G$ is finite there are finitely many classes, so steps 1.1–3.1 apply.

If $X=\mathrm{\varnothing }$, there is just the empty function, which is the zero function; there are no orbits and F4 gives the zero function as the empty sum. F5 gives the empty basis and dimension zero. With one orbit, step 2.1 reads $f=c_X{\mathbf{1}}_X$, and step 3.1 proves that this single nonzero indicator is a basis.

Pointwise addition and scaling make ${\mathbb{C}}^X$ a vector space: each abelian addition identity, both distributive identities, scalar associativity and the scalar identity hold at each $x$ by the field laws in $\mathbb{C}$. F5 defines every complex sum in the displayed form. Since $m>0$, $m^{-1}$ exists and is positive real, and F2 gives $\overline{m^{-1}}=m^{-1}$. Thus the form is defined on every pair of functions.

For complex lists $a_j,b_j$ and $c\in \mathbb{C}$, finite sums satisfy ${\sum }_j(a_j+b_j)={\sum }_j{a}_j+{\sum }_j{b}_j$, ${\sum }_{j}c{a}_j=c{\sum }_j{a}_j$ and $\overline{{\sum }_j{a}_j}={\sum }_j\overline{a_j}$. Here is the induction verifying their complex types: at length zero all sums vanish and $\bar{0}=0$. On appending $a,b$, the first formula follows by rearranging $(A+B)+(a+b)=(A+a)+(B+b)$; the second from $cA+ca=c(A+a)$; the third from $\overline{A+a}=\bar{A}+\bar{a}$. F7 proves the three identities for every length, and F5 transfers them to any finite enumeration of $X$.

For $a,b\in \mathbb{C}$ and $f_1,f_2,h\in {\mathbb{C}}^X$, expand the summand and apply step 1.2 to obtain $⟨a{f}_1+b{f}_2,h⟩=m^{-1}{\sum }_x(a{f}_1(x)+b{f}_2(x))\overline{h(x)}=a⟨f_1,h⟩+b⟨f_2,h⟩$.

Applying conjugation to the finite sum and using its involution gives $\overline{⟨h,f⟩}=m^{-1}{\sum }_x\overline{h(x)}f(x)=⟨f,h⟩$.

On the diagonal, F6 gives $⟨f,f⟩=m^{-1}{\sum }_x\mathrm{\mid }f(x){\mathrm{\mid }}^2$. These are nonnegative real summands, so F5 identifies their sum with the real finite sum of F3, and F4 shows the result is real and nonnegative. If it is zero, multiplication by $m>0$ gives ${\sum }_x\mathrm{\mid }f(x){\mathrm{\mid }}^2=0$; F4 forces each $\mathrm{\mid }f(x){\mathrm{\mid }}^2=0$, hence $f(x)=0$ by F6. Conversely $f=0$ makes every summand zero.

Steps 2.1–2.3 verify F1 and therefore give an inner product. If $X=\{x\}$ the expression is $f(x)\overline{h(x)}$ and the same verification applies. The zero function has diagonal value zero by step 2.3; the empty set is excluded precisely because the prescribed normalization would divide by zero.

Put $S={\sum }_{i=1}^n{z}_i$, using F4. For any complex finite lists, distributivity over a finite sum follows by induction: it holds for the empty sum; appending $a$ changes $cA$ to $c(A+a)=cA+ca$. Likewise conjugation through a finite sum holds at zero and is preserved by $\overline{A+a}=\bar{A}+\bar{a}$. F7 therefore gives these identities for every finite length.

Consequently $\mathrm{\mid }S{\mathrm{\mid }}^2=S\bar{S}={\sum }_{i,j}{z}_i\overline{z_j}$. Split the pairs into $i=j$, $i<j$ and $j<i$, and reindex the third part by swapping its coordinates, as permitted by F5. The diagonal sum is $n$ since $z_i\overline{z_i}=1$, giving $\mathrm{\mid }S{\mathrm{\mid }}^2=n+{\sum }_{i<j}(z_i\overline{z_j}+z_j\overline{z_i})$.

For each pair $i<j$, expand $\mathrm{\mid }{z}_i-z_j{\mathrm{\mid }}^2=(z_i-z_j)(\overline{z_i}-\overline{z_j})=2-z_i\overline{z_j}-z_j\overline{z_i}$. There are $n(n-1)\mathrm{/}2$ such pairs: the count is zero at $n=1$, and adjoining index $n+1$ adds the $n$ pairs $(i,n+1)$, so the formula is preserved by $n(n-1)\mathrm{/}2+n=n(n+1)\mathrm{/}2$. Induction gives this count for all positive $n$. Adding the displayed expansions to step 2.1 cancels each off-diagonal term and yields ${\sum }_{i<j}\mathrm{\mid }{z}_i-z_j{\mathrm{\mid }}^2=n(n-1)-(\mathrm{\mid }S{\mathrm{\mid }}^2-n)=n^2-\mathrm{\mid }S{\mathrm{\mid }}^2$.

Each squared modulus is a nonnegative real by F1 and F6. Use F2 to enumerate the pair set and F3 to conclude $n^2-\mathrm{\mid }S{\mathrm{\mid }}^2\ge 0$. Since $\mathrm{\mid }S\mathrm{\mid }\ge 0$ and $n>0$, F8 now gives $\mathrm{\mid }S\mathrm{\mid }\le n$.

If $\mathrm{\mid }S\mathrm{\mid }=n$, the identity in step 3.1 makes the sum of squared differences zero. F3 forces $\mathrm{\mid }{z}_i-z_j{\mathrm{\mid }}^2=0$ for every pair, and F6 gives $z_i=z_j$. For $n>1$, take the pairs $(1,j)$ to conclude every $z_j=z_1$. For $n=1$ that conclusion already holds and the pair sum is empty.

Conversely, if every $z_i=z_1$, repeated addition gives $S=n{z}_1$. Thus $\mathrm{\mid }S\mathrm{\mid }=\mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }{z}_1\mathrm{\mid }=n$ by F6 and the real modulus in F1. At $n=1$ this reads $\mathrm{\mid }{z}_1\mathrm{\mid }=1$. This proves both directions of the equality characterization.

Let $m_i$ be the positive order of $z_i$, and set $N=m_1\cdots m_n>0$. Each $m_i$ divides $N$, hence $z_i^N=(z_i^{m_i}{)}^{N\mathrm{/}{m}_i}=1$. The nonconstant polynomial $q(t)=t^N-1$ splits in $\mathbb{C}$ by F5. Let $R$ be its set of roots there and $E=\mathbb{Q}(R)\subset \mathbb{C}$. The linear factorization has $N$ factors, so $R$ is finite. The polynomial splits over $E$, and $E$ is generated by these roots, so it is the cyclotomic splitting field of F4. All $z_i$ belong to $E$.

Each member of $R$ is algebraic over $\mathbb{Q}$, being a root of $q\in \mathbb{Q}[t]\setminus \{0\}$. F6 makes $E\mathrm{/}\mathbb{Q}$ finite, and F7 makes it algebraic. As $1\mathrm{/}n\in \mathbb{Q}$, repeated addition and multiplication in $E$ show that $a\in E$; F10 gives the same finite sum in $E$ and $\mathbb{C}$. In particular $a$ is algebraic by F7.

F8 applies to the algebraic splitting field $E\mathrm{/}\mathbb{Q}$ of the nonzero polynomial $q$, giving normality. Let $p$ be the monic minimal polynomial of $a$ over $\mathbb{Q}$. By F12, $p$ splits over $E$. If $b\in \mathbb{C}$ is a conjugate of $a$, then $p(b)=0$ by F3. Write $p(t)={\prod }_j(t-r_j)$ with $r_j\in E$. The field $\mathbb{C}$ has no zero divisors, so ${\prod }_j(b-r_j)=0$ forces $b=r_j$ for some $j$; thus $b\in E$.

Apply F9 to the identity isomorphism of $\mathbb{Q}$ and the monic irreducible polynomial $p$, with roots $a$ and $b$. It gives a $\mathbb{Q}$-isomorphism $\tau :\mathbb{Q}(a)\to \mathbb{Q}(b)$ satisfying $\tau (a)=b$. Both fields lie in $E$. Since $\mathbb{Q}\subseteq \mathbb{Q}(a)\subseteq E$, one has $\mathbb{Q}(a)(R)=E$; the two inclusions follow respectively because $E$ contains $a,R$ and because $\mathbb{Q}(R)=E$. The same argument gives $\mathbb{Q}(b)(R)=E$. Thus $E$ is a splitting field of $q$ over each of these base fields.

The coefficients of $q$ are rational and $\tau$ fixes them, so ${\tau }_{\ast }q=q\mathrm{\ne }0$. Apply F2 to the two splitting-field structures from step 4.1. It extends $\tau$ to a field isomorphism $\sigma :E\to E$. This is a $\mathbb{Q}$-automorphism and $\sigma (a)=b$.

A field homomorphism preserves finite sums: it sends the empty sum to $0$, and $\sigma (A+z)=\sigma (A)+\sigma (z)$ preserves the assertion when a term is appended; F11 proves this for every length in F10. Since $\sigma (1\mathrm{/}n)=1\mathrm{/}n$, it follows that $b=\sigma (a)=n^{-1}{\sum }_i\sigma (z_i)$.

Multiplicativity gives $\sigma (z_i{)}^{m_i}=\sigma (z_i^{m_i})=1$. If $\sigma (z_i{)}^d=1$ for a positive $d<m_i$, applying ${\sigma }^{-1}$ gives $z_i^d=1$, contrary to the definition of $m_i$ in F1. Thus the order is exactly $m_i$. This covers $m_i=1$ and repeated roots. When $n=1$, the construction still applies with $a=z_1$. When $a=0$, its minimal polynomial is $t$, so $b=0$; steps 4.1–6.1 still apply. No step imposed integrality on $a$.

Because $I$ is finite, every tuple has finite support, so $V={\prod }_{i\in I}{V}_i$ with the operations in F3. Choose a finite basis of each $V_i$. This uses only finitely many existential witnesses. Their coordinate inclusions form a finite basis of $V$: they span because each coordinate has a basis expansion, and a zero combination has all coefficients zero by projecting to each coordinate and using that coordinate’s independence. Thus $V$ is finite-dimensional. Zero-dimensional coordinates contribute empty bases.

For $g\in G$, define $\rho (g)$ by the displayed coordinate formula. The equality ${\rho }_i(g)(a{v}_i+b{w}_i)=a{\rho }_i(g)v_i+b{\rho }_i(g)w_i$ in every coordinate proves linearity. Also ${\rho }_i(g){\rho }_i(g^{-1})={\mathrm{id}}_{V_i}$ and the reversed product is the identity, so $\rho (g^{-1})$ is a two-sided inverse of $\rho (g)$. Hence $\rho (g)\in \mathrm{GL}(V)$.

For every tuple $v$, the $i$-coordinate of $\rho (gh)v$ is ${\rho }_i(gh)v_i={\rho }_i(g){\rho }_i(h)v_i$, the $i$-coordinate of $\rho (g)\rho (h)v$. Similarly $\rho (e)v=v$. Therefore $\rho$ is a homomorphism and, with step 1.1, a finite-dimensional representation as in F1.

If $g\in \ker \rho$, F2 gives $\rho (g)={\mathrm{id}}_V$. For any $i$ and $v_i\in V_i$, apply this equality to the coordinate inclusion ${\mathrm{ȷ}}_i(v_i)$. Its $i$-coordinate yields ${\rho }_i(g)v_i=v_i$. Since $v_i$ was arbitrary, ${\rho }_i(g)={\mathrm{id}}_{V_i}$, so $g\in \ker {\rho }_i$ for every $i$.

Conversely, if $g\in \ker {\rho }_i$ for every $i$, then for every tuple $v$, $\rho (g)v=({\rho }_i(g)v_i{)}_i=(v_i{)}_i=v$. Hence $\rho (g)={\mathrm{id}}_V$ and $g\in \ker \rho$. This proves the intersection formula.

For $I=\mathrm{\varnothing }$, F3 gives $V=0$. Its unique endomorphism is its identity and is invertible, so every $g$ acts identically and $\ker \rho =G$. The condition that $g$ lie in each kernel is vacuous, giving the same $G$. For a one-element family steps 2.2–3.1 give the single coordinate kernel. For a zero coordinate its kernel is $G$, so it places no further restriction on the intersection.

If $G$ is abelian, then for all $g,h\in G$ one has $gh{g}^{-1}=hg{g}^{-1}=h$. Thus every element of ${\mathrm{Cl}}_G(h)$ is $h$; and $eh{e}^{-1}=h$ shows that $h$ belongs to the class. Hence ${\mathrm{Cl}}_G(h)=\{h\}$.

Conversely, suppose every class is a singleton. Since $eh{e}^{-1}=h$, that singleton must be $\{h\}$. For arbitrary $g,h$, F1 then gives $gh{g}^{-1}=h$. Multiplying on the right by $g$ yields $gh=hg$. Thus every pair commutes and $G$ is abelian.

In the trivial group the sole class is $\{e\}$ and $ee=ee$, so both properties hold. In all groups, the membership $h\in {\mathrm{Cl}}_G(h)$ used above prevents empty classes. The argument quantifies over arbitrary $g,h$ and makes no cardinality assumption.