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## Finite groups with at least 3 elements cannot have a subgroup consisting of all but one element

Let $G$ be a finite group with $|G| = n > 2$. Prove that $G$ cannot have a subgroup $H$ such that $|H| = n-1$.

Solution: Under these conditions, there exists a nonidentity element $x \in H$ and an element $y \notin H$. Consider the product $xy$. If $xy \in H$, then since $x^{-1} \in H$ and $H$ is a subgroup, $y \in H$, a contradiction. If $xy \notin H$, then we have $xy = y$. Thus $x = 1$, a contradiction. Thus no such subgroup exists.