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Source. The large-prime step in the Notes of proof claim 133. The cross-difference and reduced-fraction uniqueness arguments are supplied explicitly here.

For $M\ge2$, let

$$\mathcal R_M=\{(a,b):1\le a,b\le M,\ \gcd(a,b)=1\}.$$

Statement. Let $q>M^2$ be prime, and suppose that the residues $1,\ldots,M$ lie in a subgroup $H\le\mathbf F_q^*$. Then

$$\iota:\mathcal R_M\longrightarrow H,\qquad \iota(a,b)=ab^{-1}\pmod q,$$

is well defined and injective. Consequently $|H|\ge C(M)$, where $C(M)=|\mathcal R_M|$.

Complete proof. Since $b\le M<q$, its residue is nonzero and has an inverse. Both $a$ and $b$ belong to $H$, so subgroup closure gives $ab^{-1}\in H$.

Suppose $ab^{-1}=cd^{-1}$ in $\mathbf F_q$ for two members $(a,b)$ and $(c,d)$ of $\mathcal R_M$. Then $q\mid ad-bc$. Each product lies between $1$ and $M^2$, so

$$|ad-bc|\le M^2-1<q.$$

The divisibility therefore forces $ad=bc$ over the integers. Because $\gcd(a,b)=1$, the equality implies $a\mid c$; reversing the two pairs gives $c\mid a$. Positivity yields $a=c$, and then $b=d$. Thus $\iota$ is injective.

Dependencies. Elementary modular arithmetic and uniqueness of a reduced positive fraction.

Bears on. The large-prime case in the partial threshold theorem.