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 $n = |G| &gt; 2$.

How can one show this without using Lagrange's theorem (which is in chapter 3 of Dummit).

Thank you
We know that any subgroup must have the identity element in it. We also know that every subgroup must contain inverses for all of its elements. Suppose that $H$ is a subgroup of order $n-1$. Let $x$ designate the one element of $G$ not in $H$. Then $x$ must be its own inverse, as if $x^{-1} \not= x$ we have that $x^{-1}\in H$, yet $x$ is not in $H$ (which is a contradiction).

Now take any non-identity $y\in H$. Then if $xy$ is in $H$, then this implies that $x$ is in $H$ since we can multiply by $y^{-1}$. The only way that $xy$ is not in $H$ is if $y=1$. But we assumed otherwise. 

We then reach the contradiction.
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 $H$.