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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.

bluestool
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    ask yourself: can this element be identity ? if not, where will its inverse be ? – Amr Dec 01 '19 at 23:32
  • It would have to be an element of order 2 – bluestool Dec 01 '19 at 23:34
  • oh yea lol, it wasn't the most effective hint – Amr Dec 01 '19 at 23:36
  • There was already a proof of this by Ferra, in the comments to the previous question:https://math.stackexchange.com/questions/3457238/suppose-exists-a-in-g-cdot-a-neq-e-with-g-setminus-a-le-g-prove-t/3457656#comment7110643_3457656 – verret Dec 02 '19 at 01:39

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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. $\square$

user26857
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Amr
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  • I did not assume $G-{x}$ has to be a subgroup of $G$. – bluestool Dec 01 '19 at 23:44
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    @bluestool true, you didn't mention it in the question. I assumed it because otherwise the question is trivial/unnatural. Did you mean to ask that $G-{x}$ can be equipped with some group operation ? – Amr Dec 01 '19 at 23:46
  • Yes, the question is more along those lines. Why would it be unnatural then? Or trivial? – bluestool Dec 01 '19 at 23:48
  • I believe its a set theoretic fact that any (nonempty as Arturo points out) set can be equipped with a group operation. If we ignore the really large sets, then we have the following easy facts: any finite set can be equipped with a group structure , and any countably infinite set can be equipped with a group structure – Amr Dec 01 '19 at 23:50
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    @bluestool: If you are willing to define an arbitrary operation on $G-{x}$, then, assuming the Axiom of Choice, you can do it with any group except the trivial group, because every nonempty set can be given a group structure. So it is not an interesting question unless the operation you want is the “natural” one, and the only natural one is the one inherited from $G$. – Arturo Magidin Dec 01 '19 at 23:50
  • It's unnatural because in this case the group structure of $G$ itself is useless info for the question – Amr Dec 01 '19 at 23:51
  • I did not know that. Thank you! – bluestool Dec 01 '19 at 23:51
  • @Amr: “For every nonempty set $X$, there is a binary operation $\cdot$ such that $(X,\cdot)$ is a group” is equivalent to the Axiom of Choice. As to the finite case, in the absence of AC it depends on your definition of “finite”. If it means “bijectable with an element of $\omega$, then your argument works. – Arturo Magidin Dec 01 '19 at 23:52
  • @ArturoMagidin very cool, I didn't know it was equivalent to the axiom of choice. – Amr Dec 01 '19 at 23:53
  • @ArturoMagidin Yes, by finite I meant a set that is "bijectable" with some member of $\omega$. I suppose your comment implies that this won't work if one chooses to work with Dedekind finiteness , no ? – Amr Dec 01 '19 at 23:54
  • @Amr: I just mean that in the absence of AC, one has to be careful with the definitions, because some of the definitions of finite and infinite that are equivalent when AC is present may fail to be equivalent without AC. For example, “Bijectable with a proper subset” and “Not finite” are not equivalent definitions in the absence of AC. If “finite” means “bijectable with an element of $\omega$”, then it’s easy to give finite sets group structures; but if it means something else, you need to check if you can deduce that it is bijectable with an element of $\omega$ first. – Arturo Magidin Dec 01 '19 at 23:57
  • @ArturoMagidin I read your comments carefully again and got your point. Question: In absence of $AC$, can every Dedekind finite set be equipped with a group structure ? – Amr Dec 01 '19 at 23:59
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    @Amr: I honestly do not know; Asaf is probably a good person to ask. I know that Dedekind infinite implies “not bijectable with any element of $\omega$ in ZF; which means that the implication goes the wrong way to get an easy group structure on a Dedekind-finite set. – Arturo Magidin Dec 02 '19 at 00:12