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Source. The unnumbered theorem and proof on printed pp. 376–377 (PDF p. 2). This is a complete rewritten proof. Unlike Part II's initial boxes, the product sets here need not have interval projections.

Statement

Let $n\ge1$, let $p_1,\ldots,p_n$ be distinct odd primes, and let $s_i\ge1$. Let $P=\prod_{i=1}^nP_i$, where $|P_i|=p_i^{s_i}$. Consider a finite family $\mathcal T$ of nonempty proper product sets $C=\prod_iC_i\subsetneq P$ such that every $|C_i|$ is a nonnegative integer power of $p_i$.

Set

$$A_i=\sum_{r=0}^{s_i-1}p_i^r,\qquad y_i=p_i^{s_i}-A_i,\qquad x_i=\frac{A_i}{y_i} =\frac{p_i^{s_i}-1}{(p_i-2)p_i^{s_i}+1},$$ $$F(x)=\prod_{i=1}^n(1+x_i)-\sum_{i=1}^n x_i.$$

If $\mathcal T$ covers $P$ and $F(x)<2$, then two members of $\mathcal T$ have the same cardinality. The source writes $\psi(|P|)=F(x)-1$ and states the hypothesis as $\psi(|P|)<1$.

Proof

Assume all cardinalities are distinct. For a product set $C$, let

$$I(C)=\{i:C_i\ne P_i\}.$$

Since $C_i\subseteq P_i$, equality of cardinalities forces equality of these finite sets. Thus $i\in I(C)$ exactly when $|C_i|=p_i^r$ with $0\le r<s_i$. Properness makes $I(C)$ nonempty. Unique prime factorization shows that each vector of projection cardinalities occurs for at most one member of $\mathcal T$.

Let $\mathcal S$ consist of the sets with $|I(C)|=1$. Their complement

$$R=P\setminus\bigcup_{C\in\mathcal S}C=\prod_iR_i$$

is a product set. Indeed, a member with $I(C)=\{i\}$ removes just its projection $C_i$ in coordinate $i$. For that coordinate, the total size removed is at most $\sum_{r=0}^{s_i-1}p_i^r=A_i$, because each exponent occurs at most once. Hence $|R_i|\ge y_i>0$.

One may also use the source's useful bound $y_i\ge p_i^{s_i-1}$. To verify it, put $s=s_i,p=p_i$. Then

$$A_i=\frac{p^s-1}{p-1} \le(p-1)p^{s-1},$$

because $(p-1)^2\ge p$ for $p\ge3$. Thus $p^s-A_i\ge p^{s-1}$. In particular $R$ is nonempty.

For a remaining member $C$ with $I=I(C)$ of size at least two,

$$\begin{aligned} |R\cap C| &=\prod_i|R_i\cap C_i|\\ &\le\prod_{i\notin I}|R_i|\prod_{i\in I}|C_i|\\ &\le |R|\prod_{i\in I}\frac{|C_i|}{y_i}. \end{aligned} \tag{1}$$

Fix such an $I$. Summing over its possible exponent vectors, with at most one set per vector, gives

$$\sum_{C\in\mathcal T:\ I(C)=I}|R\cap C| \le |R|\prod_{i\in I} \left(\frac1{y_i}\sum_{r=0}^{s_i-1}p_i^r\right) =|R|\prod_{i\in I}x_i. \tag{2}$$

The members outside $\mathcal S$ cover $R$, since the singleton-type members miss it. Therefore

$$\begin{aligned} |R| &\le\sum_{C\in\mathcal T\setminus\mathcal S}|R\cap C|\\ &\le |R|\sum_{|I|\ge2}\prod_{i\in I}x_i =|R|(F(x)-1). \end{aligned}$$

Cancel $|R|>0$ to get $F(x)\ge2$, contradicting the hypothesis. For $n=1$ the sum over $|I|\ge2$ is empty, so this argument also excludes a cover with distinct cardinalities in that case.

Why the general product-set formulation matters

The proof used projection cardinalities, not alignment or nesting of the projections. Thus it applies to arbitrary labels of Sylow groups in the nilpotent-group corollary. For cyclic groups the more precise digit correspondence also gives the aligned boxes used in Part II.

Bears on. The first necessary condition for Problem 7.