Subgroup of a ring closed under multiplication?

Question

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.

Answer 1

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.

Answer 2

If $R=\Bbb R$ just take $G=\pi\Bbb Z$, for instance: clearly $G$ is not closed under multiplication, and $\pi\Bbb Q$ works equally well. These are of course representative of a whole family of examples.

For a completely different example, let $R=\wp(\Bbb N)$, with symmetric difference as addition and intersection as multiplication. Let

$$G=\{s\in R:s\text{ is finite and }|s|\text{ is even}\}\;;$$

it’s not hard to verify that $G$ is an additive group, but $\{1,2\}\cap\{2,3\}=\{2\}\notin G$.

Answer 3

Also In $\mathbb R$, $\{n\sqrt{m}|n\in Z\}$, m is a product of nonrepeating primes; is a subgroup under addition but not a subring.

for ex. $\{n\sqrt{2}|n\in Z\}$,$\{n\sqrt{3}|n\in Z\}$,$\{n\sqrt{6}|n\in Z\}$

Also in case of non commutative rings we have, $\{\begin{bmatrix}mi&x\\y&ni\end{bmatrix}|x,y,m,n\in Z\}$ is a subgroup under addition but not a subring.