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