For which integers $n>2$ does there exist a finite group $G$ of order $n$ such that $G$ has a subgroup of order $n-2$?

Question

For which integers $n>2$ does there exist a finite group $G$ of order $n$ such that $G$ has a subgroup of order $n-2$?

I know how to approach this problem using Lagrange Theorem, but I am not allowed to use it.

Is there a way to generate the fact "if G is a finite group and H is a subgroup of G, then |H| divides |G|" without using coset?

Answer 1

Here's how I did a similar problem before I know Lagrange's theorem.

Suppose $H$ is a proper subgroup of $G$, then there is an $x\not \in H$. It is easy to show that $hx\not\in H$ for all $h\in H$. Therefore there are at least as many elements in $G\setminus H$ as there are in $H$.

We conclude that $|H|\leq \frac{|G|}{2}$. Therefore you must have in your case that $n-2\leq n/2$ or $n\leq 4$.

Clearly it is impossible with $n=1,2$, and it is clearly possible for $n=3$, for $n=4$ you can consider the group $\mathbb Z_4$ and the subgroup $\{0,2\}$.