 Solution

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,The left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ onto that coset, explaining the formula. If $|G|=p$ is prime, then $|H|$ is either $1$ or $p$, so $H=\{e\}$ or $H=G$. Thus the correct conclusion is no proper nontrivial subgroup. The trivial subgroup $\{e\}$ is proper for every group of prime order, so the wording in the question needs this qualification.