Let $(R, +, \cdot)$ be a ring with identity $1$. Let $G \subset R$ be a group under addition. Then $G$ is a subset of $R$, so we can perform $R$'s multiplication on elements of $G$. Will multiplication in $G$ always be closed? What are some counterexamples?
If $R = \mathbb{Z}$, the only subgroups of $\mathbb{Z}$ are of the form $n \mathbb{Z}$ for nonnegative integer $n$. These are all closed under multiplication. What if we let $R = \mathbb{R}$?
EDIT: there are counterexamples when $R = \mathbb{R}$. What about when $R$ is a noncommutative ring? Also, by "closed multiplication in $G$" I mean multiplication of elements of $G \subset R$ will remain in $G$, not multiplication of elements of $G$ with elements of $R \setminus G$. I.e. $G$ should qualify as a magma with respect to $R$'s multiplication.
Best Answer
Let $R=\mathbb{C}$ the complex numbers. Then the imaginary line $\{xi|x\in \mathbb{R}\}$ is not closed under multiplication.
Similarly let $R=\mathbb{H}$ the quaternions. This is non-commutative (to answer the OP's edit). Again the imaginary line $\{xi|x\in \mathbb{R}\}$ is not closed under multiplication.