I am teaching some really advanced high school students about groups and rings and wondering of examples of groups that are not rings.

I am hoping to find such examples where addition and multiplication are actually defined for the group. I assume, for example, that the symmetric group is a group but not a ring because addition and multiplication are not defined.

Also, the cyclic subgroup of order n. Am I right that it is not a ring since it is not a group under addition? This seems clear to me, but worried about stating this in class but I know that it is isomorphic to the set of integers modulo n that DOES constitute a ring.

Any help would be appreciated!

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  • I somewhat get the spirit of your question, but maybe you could clarify what you are looking for. Trivial multiplication can be defined always, thus giving any abelian group a (non interesting) ring structure. – qualcuno Oct 07 '18 at 10:39
  • What do you mean by "cyclic subgroup of order n"? If you mean $\mathbb{Z}_n$ (integers modulo n) then it is a ring for any $n$ and even a field if $n$ is prime. Sorry, I just noticed your last comment. – badjohn Oct 07 '18 at 10:43
  • I mean the group {1,c, c^2,..., c^n} where c^n=1 and follows the usual index law c^n c^m = c^(n+m). It is a group under multiplication but not addition and I was taught that it is isomorphic to integers modulo n under addition. – Andrew Oct 07 '18 at 10:46
  • Well, one way is to just not define multiplication. Take any famous set with group properties (and maybe more) and just forget the multiplication (if any). E.g. the integers or the real numbers with addition but not multiplication. – badjohn Oct 07 '18 at 10:47
  • Guido: I don’t know what you mean by trivial multiplication. I am just wondering if there are structures that form an a group under addition, but where multiplication is defined but not associative. – Andrew Oct 07 '18 at 10:48
  • The odd integers are not closed under addition, but are closed under multiplication... – Henno Brandsma Oct 07 '18 at 10:49
  • Yes, it is isomorphic to the integers modulo $n$, just look at how the exponents behave. So, rename the operation as addition and you can add a multiplication to make it a ring. – badjohn Oct 07 '18 at 10:49
  • Here's a simple source of many examples: any non-Abelian (non-commutative) group. – badjohn Oct 07 '18 at 10:50
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    On the trivial multiplication, take any commutative group, define multiplication as $ab = 0$ for all $a$ and $b$ and you have a ring (unless you require a ring to have a multiplicative identity, some do and some don't). – badjohn Oct 07 '18 at 10:52
  • For multiplication defined but not associative, I think that someone mentioned the octonians but I cannot see it now. https://en.m.wikipedia.org/wiki/Octonion Also, there are vectors in 3 or 7 dimensions with the cross-product. https://en.m.wikipedia.org/wiki/Cross_product. There is a connection between these examples. – badjohn Oct 07 '18 at 11:05

3 Answers3


As is common, the answer depends on exactly what you allow.

Groups are not required to be Abelian (commutative) but addition in a ring is required to be commutative. Hence no non-Abelian group can be extended to a ring.

So, from now onwards, let's assume only Abelian groups.

Here's one source of trivial examples. Take any ring or field (e.g. the integers or the real numbers) and just forget the multiplication. E.g. define the unreal numbers as like the real numbers but multiplication is not defined. They are a group under addition but not a ring.

Let's ban those examples.

Are there any natural examples of structures which are groups under addition, have a multiplication defined, but are not rings? Yes. Guido gives one example: The Octonians (Wikipedia). Another example, one that is more commonly used, are vectors in $\mathbb{R}^3$ with the cross product (Wikipedia). This is not associative and hence not a ring.

Can we take any Abelian group and extend it to a ring? Some people insist that a ring must have a multiplicative identity (unity) and some don't. If you don't require a ring to have a unity then the trivial multiplication is always possible.

$\forall x, y : xy = 0$

Let's require a unity. In this case, some interesting examples fail to be rings. Infinite matrices with only finitely many non-zero coefficients; integrable functions from $\mathbb{R}$ to $\mathbb{R}$.

However, that one multiplication fails to be a ring does not prove that no other multiplication might work. Can any Abelian group be extended to a ring? I am not sure here but many can be. If the group is finite or finitely generated then it is isomorphic to a direct sum of finitely many copies of $\mathbb{Z}$ and various $\mathbb{Z}_n$. Multiplication can then be defined componentwise. See Abelian groups (Wikipedia)

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Let me quote the Wikipedia,

A group is a set, G, together with an operation $\cdot$ (called the group law of G) that combines any two elements $a$ and $b$ to form another element, denoted $a\cdot b$ or $ab$. To qualify as a group, the set and operation, (G, $\cdot$), must satisfy four requirements known as the group axioms: ....

Notice how they refer to a group as (G, $\cdot$). This emphasizes the fact that a group is precisely two things: A set G and an operation $\cdot$ which combines two elements of G together to form another element of the set G.

So, here are some examples of groups: $(\{0\}, +)$, $(\mathbb Z, +)$, $(\mathbb R, +)$, $(\mathbb R^n, +)$, $(\mathbb R\setminus\{0\}, \cdot)$, $(\mathbb R^+, \cdot)$ where $\mathbb R^+$is the set of positive real numbers, modular arithmetic $(\mathbb Z_n, +)$, $(\mathbb Z_p, \cdot)$ where $p$ is prime, the symmetry group of an equilateral triangle, the symmetry group of a square, and the Rubik's Cube Group. Notice that the group operation does not need to be denoted as a +. You can use any operation that takes two elements of G and maps them to G as long as that operation obeys the group axioms: closure, associativity, existence of an inverse, and an identity element.
$$ $$ A Ring, on the other hand, is three things (R, +, $\cdot$) where R is a set, and + and $\cdot$ are two operations that combine two elements of R to form another element of R. As with groups, there are certain laws that the operations have to obey, for example the distributive law, the associative law, the operation + needs to obey the commutative law, ....

As you stated, you need to have a multiplication operation to form a Ring, so for example $(\mathbb Z, +)$ is not a Ring, but $(\mathbb Z, +, \cdot)$ is a Ring.

As you noticed, many Groups have a natural extension to Rings:

  • The Group $(\mathbb Z, +)$ admits normal integer multiplication to form the ring $(\mathbb Z, +, \cdot)$.
  • The Group $(\mathbb R, +)$ also admits multiplication to form the ring $(\mathbb R, +, \cdot)$.
  • The Group $(\mathbb Z_n, +)$ admits modular multiplication to form the ring $(\mathbb Z_n, +, \cdot)$.
  • The Group $(\mathbb R^+, \cdot)$ admits a multiplication operator to form the ring $(\mathbb R^+, \oplus, \odot)$ with $a\oplus b= a\cdot b$ and $a\odot b= \exp(\log(a)\log(b))$.

Some Groups do not admit a Ring structure. For details, see e.g. this post.

$(\mathbb R^3, +, \times)$ where $u\times v$ is the cross product of $u$ and $v$ is almost a ring, but it has no identity element for multiplication and $\times$ is not associative.

If the $\cdot$ in $(R,+,\cdot)$ is associative without an identity, then $(R,+,\cdot)$ is not a Ring. It is referred to as a Rng. Here are two nice examples from the Wikipedia:

A simple example of a rng that is not a ring is given by the even integers with the ordinary addition and multiplication of integers. Another example is given by the set of all 3-by-3 real matrices whose bottom row is zero.

$$ $$ You wrote "Also, the cyclic subgroup of order n. Am I right that it is not a ring since it is not a group under addition? This seems clear to me, but worried about stating this in class but I know that it is isomorphic to the set of integers modulo n that DOES constitute a ring.".

Every cyclic group is isomorphic to either $(Z, +)$ or $(Z_n, +)$. When you say the cyclic subgroup of order $n$, perhaps you are thinking of $(\{g^0, g^1, g^2, \ldots, g^{n-1}\}, \cdot)$ with $g^a\cdot g^b= g^{\mathrm{mod}(a+b,n)}$. You might think that this group cannot be extended to a Ring because the + operator is not defined. It can be extended to the Ring $(\{g^0, g^1, g^2, \ldots, g^{n-1}\}, \oplus, \odot)$ where $g^i\oplus g^j=g^{\mathrm{mod}(i+j,n)}$ and $g^i\odot g^j=g^{\mathrm{mod}(i\cdot j,n)}$.

If you have further questions, maybe you could put them in a comment below.

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  • Not everyone requires that a ring has a multiplicative identity. I prefer not to. So, I never use rng but I say "ring with unity" or "unital ring" when required. This disagreement has persisted for as long as I know but I first saw rng only recently. – badjohn Oct 07 '18 at 19:30

The octonions are a group with addition, but the product fails to be associative.

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