**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.1 Exercise 2.1.5**

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.