Group Theory – Subgroup of Order $n-1$ of a Group of Order $n$

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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

Best Answer

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.

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