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An axis-parallel box in ${\mathbb{R}}^n$ is a Cartesian product $I_1\times \cdots \times I_n$ of bounded intervals. Its volume is the product of its side lengths, and each coordinate projection is the product of the corresponding intervals.

For four real-valued functions on a finite Cartesian product, let $\Lambda (f_{00},f_{01},f_{10},f_{11})={\mathbb{E}}_{x_0,x_1,y_0,y_1}{\prod }_{i,j=0}^1{f}_{ij}(x_i,y_j)$. Repeated Cauchy-Schwarz inequalities giveFirst separate the two $y$ averages and apply Cauchy-Schwarz inequality in $(x_0,x_1)$. Each resulting squared factor is ${\mathbb{E}}_{y,y^{\prime }}({\mathbb{E}}_{x}f(x,y)f(x,y^{\prime }))({\mathbb{E}}_{x}g(x,y)g(x,y^{\prime }))$, bounded by $\Vert f{\Vert }_{\square }^2\Vert g{\Vert }_{\square }^2$ by another Cauchy-Schwarz inequality. Expanding the four factors of $f+g$ and applying this inequality to each of the sixteen terms gives the triangle inequality for the box norm.

For nonempty finite sets $X,Y$ and a real-valued function $f$ on their Cartesian product, using uniform expectations, defineThis equals ${\mathbb{E}}_{y,y^{\prime }}({\mathbb{E}}_{x}f(x,y)f(x,y^{\prime }){)}^2$. Hence it is nonnegative, and vanishing forces $f=0$ by taking $y=y^{\prime }$. Absolute homogeneity and the box Cauchy-Schwarz inequality establish that it is a norm. This formula concerns real-valued functions; complex functions require appropriate complex conjugates in the definition.

For a real-valued function $H$ on a nonempty finite Cartesian product, let $T_{H}v(x)={\mathbb{E}}_{y}H(x,y)v(y)$ with uniform inner products. The box norm obeys $\Vert H{\Vert }_{\square }^4=\mathrm{Tr}((T_H{T}_H^{\ast }{)}^2)={\sum }_j{\sigma }_j^4$, where ${\sigma }_j$ are its singular values. For a biregular graph this separates the constant component ${\gamma }^4$ from the other fourth powers.

Every Cartesian product of cap sets is a cap set. In characteristic three, $x+y+z=0$ with any two equal forces all three equal. A zero-sum triple in a Cartesian product of cap sets is therefore diagonal in every coordinate block, hence diagonal overall.

For a real-valued function $H$ on a nonempty finite Cartesian product $X\times Y$, its normalized cut norm is the maximum of $|{\mathbb{E}}_{x,y}H(x,y)1_A(x)1_B(y)|$ over $A\subseteq X,B\subseteq Y$. It measures all rectangular subset discrepancies. The convention using arbitrary test functions bounded by one is equivalent within a factor of four, by decomposing their positive and negative parts into indicator functions.

Definiteness says that a norm vanishes only at the zero vector. The box norm on a nonempty finite Cartesian product is definite because its defining sum of squares includes equal-column terms, each of which is the square of an average of squares. Omitting repeated coordinates would destroy this proof and can produce a different quantity.

For a coordinate projection map onto $X_i$, write $X_j$ for the other factor. The map is surjective exactly when $X_i$ is empty or $X_j$ is nonempty. It is injective exactly when $X_i$ is empty or $X_j$ has at most one element. In particular, a surjection onto an empty Cartesian product need not induce a surjection onto a nonempty factor.

A finite Cartesian product of countable sets is countable. For a fixed finite exponent $m$, index tuples in ${\mathbb{N}}^m$ can be listed by increasing maximum coordinate, with each group finite. If $B$ is countably infinite and $m>0$, its power $B^m$ is also countably infinite.

A useful finite conjunction of closure assertions requires empty set, pairing, set union, set difference, Cartesian products, all finite tuple domains $A^n$, and the following operations on their relations: relative complements, intersections, coordinate projections and pullbacks along finite coordinate maps, together with equality and membership restricted to $A^2$. The arities and maps are quantified objects, so these requirements form finitely many first-order formulas, not an infinite schema. In a transitive set, individual finite tuples exist by pairing and union, making the tuple-domain assertions correct externally. Atomic truth relations, Boolean operations and projection then compute the truth table of each coded first-order formula correctly by finite induction. Thus satisfaction for a set structure and the relation $B=\mathrm{Def}(A)$ are absolute for shared arguments. Every nonzero limit level of the relative constructible hierarchy is closed under these operations, since each result can be defined at finitely many subsequent stages.

A finite tuple is an ordered list of finitely many entries; coordinates can repeat, and their positions matter. A tuple of length $n$ from a set $M$ is an element of the Cartesian product $M^n$. The empty tuple has length zero. Finite tuples support Scott formulas in a back-and-forth method construction, where matching tuples determine a partial isomorphism of structures.

Work with $x$ modulo $p-1$ and $u$ modulo $p$. By Fermat's little theorem, $a^{p-1}=1$ in ${\mathbb{F}}_p$, so the factor $a^y$ is independent of the chosen representative of $y$. The proposed multiplication is therefore well defined on the stated Cartesian product. Its two associative bracketings have the same second component:and both have first component $x+y+z$. The identity element is $(0,0)$, and the two-sided inverse element isThus the multiplication defines the twisted cyclic pair group, of order $p(p-1)$.

If $a=1$, multiplication is coordinatewise addition, so the group is abelian. If $a\ne 1$, then $p>2$ and both $(1,0)$ and $(0,1)$ are available; their products in opposite orders are $(1,1)$ and $(1,a)$. They differ. HenceBoth $H$ and $K$ contain the identity element and are closed under the multiplication and inverses: they are the cyclic groups of orders $p-1$ and $p$ respectively. Conjugation gives, with all coordinates reduced in the appropriate modulus,so $K$ is a normal subgroup. For $H$, the corresponding calculation isWhen $a=1$ this lies in $H$. Conversely, for $a\ne 1$, take $x=0$, $u=1$, $y=1$: the second coordinate is $a-1\ne 0$, so $H$ is not normal. Thus $H$ is normal precisely when $a=1$.

Finally, the projectionis a surjective group homomorphism because the first coordinates add. Its kernel of a group homomorphism is exactly $K$. At $p=2$, $a=1$ is the only possibility and the first factor is trivial, consistent with every conclusion. No assumption that $a$ generates the multiplicative group of the finite field is needed.