"Rings are groups"

I've read in many places that "rings are groups", for example:

Formally, a ring is an abelian group whose operation is called addition, with a second binary operation called multiplication that is…

I've also read that a group is a set of elements and a (one) binary operator over that set, for example on Wikipedia (first paragraph):

"Groups are one binary operator over a set"

In mathematics, a group is a set equipped with an operation that…

  • And again in the same article, in the more detailed section #Definition:

A group is a set $G$ together with a binary operation on $G$, here denoted "$\cdot$", that…


Obviously these are inconsistent because a group is supposed to have one operator based on these definitions above. I need more consistent definitions for "ring" and "group". Thank you in advance.

The following un-official definitions provide a more consistent alternative, but the problem with them is that I made them up. I need generally-accepted consistent definitions.

  • group: a set of elements and at least one binary operator(s) over that set
  • ring: a group with exactly two operators: addition and multiplication

Please tell me if I'm missing something. I feel like I am.

  • 8
    $\begingroup$ If (R, +, *) is a ring with the two binary operation +, *, then the set R with the binary operation + is a group i.e. (R, +) is a group. It's that simple really. Now the definition of group that I'm familiar with: is that a group has one operation. If another text defines a group differently, then that's another matter. $\endgroup$
    – I0_0I
    May 23, 2021 at 9:14
  • 3
    $\begingroup$ There is no inconsistency, since a group has one binary operation, and a ring has two. But $1<2$, so the number of binary operations is no contradiction. $\endgroup$ May 23, 2021 at 9:29
  • 1
    $\begingroup$ No, $(R,*)$ is not an abelian group except when $R = \{0\}$. $\endgroup$
    – Derek Holt
    May 23, 2021 at 14:41
  • 6
    $\begingroup$ I don't really understand the closure here. There is a conceptual confusion which had led to a genuine question, which is allowed, right? $\endgroup$
    – user1729
    May 23, 2021 at 17:40
  • 3
    $\begingroup$ The closure, deletion and downvotes on this question are all quite disturbing. Remember, this site has always welcomed math questions at all levels. $\endgroup$ May 23, 2021 at 21:46

2 Answers 2


If you want to do things very formally, then a group $G$ is a pair $G=(G_\text{set},\star)$, where $G_\text{set}$ is a set and $\star\colon G_\text{set}\times G_\text{set}\to G_\text{set}$ is an operation such that certain axioms are satisfied. Similarly, a ring $R$ is a triple $R=(R_\text{set},+,\times)$ where $R_\text{set}$ is a set, both $+$ and $\times$ are operations on $R_\text{set}$ and certain axioms are satisfied. One of these axioms is that the pair $(R_\text{set},+)$ is an abelian group.

However, mathematicians are lazy and language needs to be efficient, so usually we will see a group $G=(G_\text{set},\star)$ being treated as a set itself and we write statements like $g\in G$ when formally it should be $g\in G_\text{set}$. But everybody knows which set is meant, so we drop the distinction between $G_\text{set}$ and $G$. In this way, a group $G$ is a set with some additional structure (the operation $\star$).

Similarly, when $R=(R_\text{set},+,\times)$ is a ring, you always have the group $(R_\text{set},+)$ and we may say that a ring is a group with some additional structure (the operation $\times$). Now the symbol $R$ is used to denote all three: the underlying set, the abelian group and the ring.

  • $\begingroup$ Wouldn't it be better to say $R = (R_{set}, (+, ×))$? Because even though $(R_{set}, +)$ is an abelian group, so is $(R_{set}, ×)$. Frankly it would make more sense if we said "every ring has two groups" rather than saying "every ring is one group". $\endgroup$
    – SMMH
    May 23, 2021 at 12:14
  • 1
    $\begingroup$ @SMMH As I said in an earlier comment, it is not true that $(R_{\rm set},\times)$ is a group except when $R_{\rm set} = \{0\}$. $\endgroup$
    – Derek Holt
    May 23, 2021 at 14:44
  • $\begingroup$ @SMMH The pair $(x,(y,z))$ encodes the exact same data as the triple $(x,y,z)$, so it doesn't matter which way you define it, as long as you stick to one definition. This is another reason to just not use such a formal definition: it requires irrelevant choices. Saying "a set with two operations" without mentioning how to encode that data is enough to get everybody on the same page. $\endgroup$
    – Christoph
    May 23, 2021 at 17:21

Actually, I think what you are finding odd is that when a group has only one binary operation how can a ring be a group because it has 2 operations. The answer is that a set with a operation is considered a group, not the set only. A set can be a group with two operations simultaneously. A ring is a set that forms an abelian group with respect to one operation and follow some other properties with another Operation if we are able to define. Concluding, All rings are abelian group with an operation, I want you to mark that it says an operation not one operation.

  • $\begingroup$ Regarding your last sentence: "an operation" literally means "one operation". $\endgroup$
    – SMMH
    May 23, 2021 at 12:09
  • $\begingroup$ An is an indefinite article meaning one or one more.👍 $\endgroup$ May 23, 2021 at 22:37
  • 1
    $\begingroup$ @VenkateshBarnwal No, the word “an” used preceding a noun refers to exactly one of that noun and certainly never “or more than one.” ‘Indefinite’ means that it does not necessary refer to a particular thing, and it has nothing to do with the quantity being indefinite. But it is true that the statements “this has two foos” and “this has a foo” are not mutually exclusive, which seems to be the OPs misconception. $\endgroup$
    – rschwieb
    May 24, 2021 at 3:40

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