Subgroup of order $n-1$ of a group of order $n$ Here is question 2.1.5 from Dummit and Foote : Prove that $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G| > 2$.
How can one show this without using Lagrange's theorem (which is in chapter 3 of Dummit).
Thank you

Consider an element $x$ in $G$ and not in $H$. Since $|H| = n - 1$, $x^{-1}$ must be in $H$. However, the inverse is unique so it's easy to see that $x^{-1}$ has no inverse in $H$ because $x$ is not in $H$.
