Let $G$ be a group such that there exists $a\in G$ such that $H = G \setminus\{a\}$ is a subgroup of $G$. Show that $|G|=2$.
This problem was given as my homework in Abstract Algebra. But I don't seem to find a way to prove this without using Lagrange's Theorem (order of subgroup must divide the order of the group that the subgroup is in) because I haven't learned that yet in class.
I also don't know if I should assume $G$ is finite or cyclic in order to show $|G|=2.$ Should I first show that the group $G$ is finite first?
Please help me on how I should approach this problem.
You can approach it by thinking about the consequences of putting a hole in the multiplication table: assume $H = G \setminus \{a\}$ is a subgroup of $G$ and let $b \in H$. I claim that $ab^{-1} \not\in H$, for if $ab^{-1} \in H$, then as $b \in H$ and $H$ is a subgroup, then $a = (ab^{-1})b \in H$, contradicting the definition of $H$ as $G \setminus \{a\}$. So we must have $ab^{-1} = a$, implying that $b = 1$. So $H = \{1\}$ and $G = \{1, a\}$.