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
 A: 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.
A: Take a look at $gH$, where $g \in G$, $g \notin H$.  Then I claim $gH \cap H = \varnothing$; for if $h \in gH \cap H$, then we have $h \in H$ and also $h \in gH$, whence $h = gk$ for $k \in H$.  Then $g = hk^{-1} \in H$,  a contradiction.  So  $gH \cap H = \varnothing$; but then
since $gH$ and $H$ each have $n - 1$ elements, $H \cup gH$ has $2(n - 1) = 2n - 2 > n$ elements if $n > 2$.  This contradicts $\mid G \mid = n$.  So such a subgroup $H$ with $\mid H \mid = n - 1$ cannot exist.  QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Let $g$ be an element of $G$ that is not in $H$.  Let $h$ be any nonidentity element of $H$.  Now show that $gh$ is not in $H$.  
A: 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 $H$.
