= Solution
For a subgroup $H$ of a finite group $G$, <Lagrange's theorem> states
$$
\boxed{|G|=[G:H]|H|.}
$$
The 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 \b[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.
Back to article page
