# Subgroup of a ring closed under multiplication?

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.

• $\mathbb Z\subseteq \mathbb R$, but $2.1\cdot 1\notin \mathbb Z$ Jul 19, 2020 at 4:31
• Yes, but $2.1 \notin \mathbb{Z}$. Multiplication of elements in $\mathbb{Z}$ will be closed. Jul 19, 2020 at 4:55
• $\frac 1 2 ℤ ⊆ ℚ$. Jul 19, 2020 at 4:57
• @RiversMcForge The questioner means $GG ⊆ G$ by “multiplicatively closed”, as is clear from his or her objection to Kentas attempt at a counterexample and which is also the usual meaning of “multiplicatively closed”. He or she’s looking for subrngs, not ideals. Jul 19, 2020 at 5:03
• For non-commutative non-division rings, consider $⟨\left[\begin{smallmatrix}0 & 1 \\ 1 & 0\end{smallmatrix}\right]⟩$ in $\operatorname{Mat}_{2×2} R$ for your favorite ring $R$ with $R ≠ 0$. Jul 19, 2020 at 5:27

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.

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 $$\{1,2\}\cap\{2,3\}=\{2\}\notin G$$.

• The answer is fine, of course, but the examples are still somewhat complicated, and unneccessarily so. Jul 19, 2020 at 5:05
• @k.stm: Matter of opinion and background: I consider the first one extremely elementary and not at all complicated. The second is pretty natural, I think, for someone like me whose background is much more in set theory than in abstract algebra. I am much more familiar with that ring than with the quaternions, for instance! Jul 19, 2020 at 5:16
• I like the $R = 2^{\mathbb{N}}$ example because it seems more creative than using fields or skew fields. (Although I haven't yet verified that $\{ f: \mathbb{N} \to \mathbb{Z}_2 \}$ under normal function addition and multiplication is not a field or skew field.) Jul 19, 2020 at 5:17
• @BrianM.Scott Both are not complicated, but both are not as visual and immediately clear as for instance the imaginary line within the complex numbers. It’s all I meant. Jul 19, 2020 at 5:23
• @k.stm: I repeat: that depends on the person. For me my first example is at least as immediately clear as the example of the imaginary line, and I don’t consider any of the posted examples to be visual. Jul 19, 2020 at 16:10

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.

• Can you prove $\{ a b m : a, b \in \mathbb{Z} \}$ contains no elements from $\{ n \sqrt{m} : n \in \mathbb{Z} \}$? I wonder if the "$m$ is squarefree" requirement could be made more general to include other integers. Jul 19, 2020 at 14:38
• Would this do the trick? Let $m$ be any integer such that $\sqrt{m}$ is not an integer, so $\sqrt{m}$ is irrational. Then $abm$ is always rational, but $n \sqrt{m}$ is always irrational. Jul 19, 2020 at 14:42
• @jskattt797, yes you can do that, that will be subgroup of one of the groups I mentioned above, and you can apply the same logic to examples given by others also to your question. Jul 20, 2020 at 17:02
• I'm talking about groups of the form $\sqrt{m}\mathbb{Z}$ for $m \in \mathbb{Z}$ such that $\sqrt{m} \notin \mathbb{Z}$. Any squarefree integer $m$ will satisfy this property, as well as many other integers, e.g. $8$. Thus, such groups generalize the $\sqrt{m} \mathbb{Z}$ (where $m$ is squarefree) example you provided. What do you mean by "that will be a subgroup"? Jul 21, 2020 at 3:03
• @jskattt797 I meant to say for ex. $n\sqrt{12}$ is a subgroup of $n\sqrt{3}$. Jul 21, 2020 at 5:01