Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 8 / ii / Solution

For a subgroup $H$ of a finite group $G$, Lagrange's theorem statesThe left cosets partition the group and each has $|H|$ members, proving the formula. If $|G|=p$ is prime, the only divisors are one and $p$, so $H=\{e\}$ or $H=G$. Hence a group of prime order has no nontrivial proper subgroups.

The word “nontrivial” is necessary: $\{e\}$ is a proper subgroup of every prime-order group. Thus the source's shorter wording cannot literally exclude all proper subgroups.