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.
from Hot Weekly Questions - Mathematics Stack Exchange
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