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Consider a group $(G,\cdot)$ with the property that $\exists a\in G, a\neq e$, such that $G\setminus \{a\}$ is a subgroup of $G$. Prove that $(G,\cdot) \cong (\mathbb{Z}/2\mathbb Z,+)$.

We know that if $H$ is a subgroup of $G$ then $\forall x \in H, y\in G\setminus H$ we have that $xy \in G \setminus H$.
In our case, $\forall x\neq a$, $xa \in G \setminus (G\setminus \{a\})=\{a\}$ and this implies that $\forall x\neq a$, $x=e$.
As a result, $G=\{e,a\}$ and it is well-known and easy to prove that any group of order $2$ is isomorphic to $(\mathbb{Z}/2\mathbb Z,+)$.

I would like to know if my proof is correct.

Shaun
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Alexdanut
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  • looks correct! If you want another proof, it follows immediately from Lagrange's theorem. – Ferra Nov 30 '19 at 17:54
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    @Ferra Thank you ! However, it doesn't follow from Lagrange's theorem because $G$ is not finite. – Alexdanut Nov 30 '19 at 17:54
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    Yes you're right. Then you can see it this way: $G$ can be written as a disjoint union of cosets, all of which are in bijection with $H$. Thus if $G\setminus {a}$ is a subgroup, then ${a}$ is a coset, thus all cosets contain exactly one element, and the claim follows. – Ferra Nov 30 '19 at 18:07

1 Answers1

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Let $|G|=m$. The order of $G-\{a\}$ is $m-1$. If this set is a subgroup, then by Lagrange's theorem, $m-1\vert m$. The only integer $m$ satisfying this condition is 2, so $|G|=2$.

The only group of order 2, up to isomorphism, is $\left(\mathbb{Z}/2\mathbb{Z},+\right)$, hence the result.

bluestool
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  • You are assuming that $G$ is finite (because of the hypothesis of Lagrange's Theorem), right? – manooooh Nov 30 '19 at 23:18
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    Yes, since the condition that removing an element yields a subgroup does not make sense for infinite groups, as far as I know. – bluestool Nov 30 '19 at 23:20
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    @bluestool why wouldn't it make sense for infinite groups? – Alexdanut Nov 30 '19 at 23:41
  • Because removing 1 element from a group removes closure under products if the group is infinite – bluestool Dec 01 '19 at 00:09
  • @bluestool Could you prove this? I don't see why this holds for any infinite group. – Alexdanut Dec 01 '19 at 22:23
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    I asked a question to get a rigorous argument as to why this is the case. I'm almost certain it should be true, and I can support it heuristically, but I cannot think of a rigorous argument to prove it. – bluestool Dec 01 '19 at 23:34