I have been challenged to give a rigorous answer to the question:
Can removing a single element from an infinite group still yield a group?
Intuitively, I would expect that removing a single element from a group $G$ would make closure under products fail. For instance, removing $6\in\mathbb{Q}^\times$ negates closure, as $2\cdot3$ is no longer in the group. However, a formal statement to prove this does not seem obvious to me.
In the finite group case, the answer is yes, as $\mathbb{Z}/2\mathbb{Z}$ satisfies this condition (removing 1 yields the trivial group).
Can anyone provide a rigorous argument as to why the answer to this question is no (or yes)?
Thank you.
Let $x$ be the element that is going to be removed. Pick $y\in G\setminus\{e,x\}$. $xy\not=x$ as $y\not = e$. Since $xy,y\in G\setminus\{x\}$ and $G\setminus\{x\}$ is subgroup, thus:
$$(xy)y^{-1}\in G\setminus\{x\}$$ $$x\in G\setminus\{x\}$$ Contradiction. $\square$