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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.

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    $\begingroup$ $\mathbb Z\subseteq \mathbb R$, but $2.1\cdot 1\notin \mathbb Z$ $\endgroup$
    – Kenta S
    Jul 19, 2020 at 4:31
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    $\begingroup$ Yes, but $2.1 \notin \mathbb{Z}$. Multiplication of elements in $\mathbb{Z}$ will be closed. $\endgroup$
    – jskattt797
    Jul 19, 2020 at 4:55
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    $\begingroup$ $\frac 1 2 ℤ ⊆ ℚ$. $\endgroup$
    – k.stm
    Jul 19, 2020 at 4:57
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    $\begingroup$ @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. $\endgroup$
    – k.stm
    Jul 19, 2020 at 5:03
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    $\begingroup$ 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 $R ≠ 0$. $\endgroup$
    – k.stm
    Jul 19, 2020 at 5:27

3 Answers 3

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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.

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

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  • $\begingroup$ The answer is fine, of course, but the examples are still somewhat complicated, and unneccessarily so. $\endgroup$
    – k.stm
    Jul 19, 2020 at 5:05
  • $\begingroup$ @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! $\endgroup$ Jul 19, 2020 at 5:16
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    $\begingroup$ 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.) $\endgroup$
    – jskattt797
    Jul 19, 2020 at 5:17
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    $\begingroup$ @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. $\endgroup$
    – k.stm
    Jul 19, 2020 at 5:23
  • $\begingroup$ @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. $\endgroup$ Jul 19, 2020 at 16:10
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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.

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  • $\begingroup$ 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. $\endgroup$
    – jskattt797
    Jul 19, 2020 at 14:38
  • $\begingroup$ 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. $\endgroup$
    – jskattt797
    Jul 19, 2020 at 14:42
  • $\begingroup$ @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. $\endgroup$
    – amitava
    Jul 20, 2020 at 17:02
  • $\begingroup$ 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"? $\endgroup$
    – jskattt797
    Jul 21, 2020 at 3:03
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    $\begingroup$ @jskattt797 I meant to say for ex. $n\sqrt{12}$ is a subgroup of $n\sqrt{3}$. $\endgroup$
    – amitava
    Jul 21, 2020 at 5:01

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