# Why is the commutator defined differently for groups and rings?

The commutator of two elements in a group is defined as $[g, h] = g^{−1}h^{−1}gh.$ In a ring, the commutator of two elements is $[a, b] = ab - ba.$

I'm asking because a ring is a (abelian) group under addition, so I would have expected it to be $[g, h] = -g-h+g+h.$

• If we took that as the commutator in a ring, then every commutator would be zero. – Omnomnomnom Jun 29 '15 at 21:44
• A group can be phrased in terms of multiplicative notation or additive. When we use additive notation it is usually because we know that it is commutative. The ring operation + is always commutative so the commutator is not very interesting, so when they talk about the commutator of a ring they are talking about the multiplicative operation. – Gregory Grant Jun 29 '15 at 21:45
• here we have two comments, each of which could serve as a distinct and worthwhile answer. Funny. – James S. Cook Jun 29 '15 at 21:50
• @JamesS.Cook I have a longer answer in the works fwiw – Omnomnomnom Jun 29 '15 at 21:51
• Yeah, I wish there was some smaller rep associated with comments in instances like these. I think good comments like yours here deserve some internet points... – James S. Cook Jun 29 '15 at 22:58

The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the "extent" to which two elements fail to commute. That is, given elements $a,b$, we wish to "compare" $ab$ and $ba$.

In the context of a group, we only have one operation: "multiplication". One way to compare two "values" using multiplication is to divide them; the "closer" you are to the identity, the closer the two values are. In particular, if $a$ and $b$ commute, we'd have $$\frac{ab}{ba} = 1$$ Where $1$ denotes the group identity here. Of course, this "division" notation is a bit ambiguous in the context of groups, since multiplication is non-commutative here. What we take to be the commutator, then, is $$[a,b] = (ba)^{-1}(ab) = a^{-1}b^{-1}ab$$ With rings, it would be fantastic if we could reuse the same definition. However, the problem is that in a ring, we can't divide. Rings don't need to be groups under multiplication, they only need to be monoids (or semigroups, depending on definition).

We still, however, have an operation that allows us to "compare" $ab$ and $ba$, namely, subtraction. In particular, when $a$ and $b$ commute, we have $ab - ba = 0$. So, because it's the only way to "measure" the extent to which $a$ and $b$ commute, we define $$[a,b] = ab - ba$$ in rings. It so happens that if your ring happens to allow division (that is, if you have a division ring), then it would seem that you could use both commutators. However, the group commutator works for every group, and the ring commutator works for every ring. This is why we define them the way we do.

• @MartinBrandenburg it is a slight (and I hope helpful) abuse of terminology, in the hopes of making things more intuitive. Perhaps it is better to say that "we have one binary operation, which I refer to as multiplication" – Omnomnomnom Jun 29 '15 at 22:10
• @MartinBrandenburg, it is not wrong: it is only if you want it to be. One can very well define a group to be a set with one operation satisfying certain conditions, and that is standard practice. – Mariano Suárez-Álvarez Jun 29 '15 at 23:48
• OK, so suppose we do consider a division ring. The usual example is $\mathbb{H}$, the quaternions. What is the commutator of $i$ and $j$ in $\mathbb{H}$? If we use the ring structure, we get $[i,j] = ij - ji = k + k = 2k$. If instead we use only the multiplicative group structure, we have $[i,j] = i^{-1}j^{-1}ij = (-i)(-j)ij = (ij)(ij) = k^2 = -1$. And of course if we take quaternions $a,b$ that actually commute, then $[a,b]$ in the ring sense is zero, while in the multiplicative sense the commutator is one. – Jeppe Stig Nielsen Jun 30 '15 at 12:17
• The last part is a bit off. Even in a division ring, we cannot use the group version of the commutator, as it would not be defined on all elements (or at least we would need to limit the definition, which is not necessary with the version for rings. – Tobias Kildetoft Jun 30 '15 at 19:16
• I didn't mean that the two commutators are the same on a division ring, only that a division ring has an underlying group, to which we could apply the "group" commutator. – Omnomnomnom Jun 30 '15 at 20:08

Here is a more abstract and unified approach to commutators. I could phrase this whole answer in terms of category theory (and this is probably the best way to think of it), but I won't in order to make the answer more accessible.

Let $G$ be a group. We want to make it commutative. What does that mean? We want to find a commutative group $A$ and a homomorphism $\pi : G \to A$ which is the "best" choice of a homomorphism to a commutative group. This entails in particular that $\pi(gh)=\pi(hg)$ and hence $\pi(ghg^{-1} h^{-1})=1$, i.e. the elements of the form $ghg^{-1} h^{-1}$ lie in the kernel of $\pi$. So the best choice should be: We let $G'$ be the (automatically normal) subgroup of $G$ generated by these elements, and let $A=G/G'$, and of course $\pi$ is the projection.

Now let $R$ be a ring. We want to make it commutative, i.e. we want to find a commutative ring $A$ with a "best" homomorphism $\pi : R \to A$. This entails in particular $\pi(ab)=\pi(ba)$ and hence $\pi(ab-ba)=0$. So we should define $A=R/I$, where $I$ is the ideal generated by elements of the form $ab-ba$.

In each case, commutativity was achieved by setting certain elements to the identity element for the group operation, which are therefore called commutators: They measure the failure of commutativity.

However, we are not limited to groups or rings. A very similar algebraic structure is a monoid, which is a semigroup with identity. Now given a monoid $M$, in order to find a commutative monoid $A$ equipped with a "best" homomorphism $\pi : M \to A$, we again have $\pi(ab)=\pi(ba)$, but now we are stuck: We cannot simplify this. But that's ok, we may just define $\sim$ to be the smallest congruence relation on $M$ which satisfies $ab \sim ba$. It is a bit ugly to describe this explicitly. But, being a congruence relation, we may construct a monoid $M/{\sim}$, which is commutative by construction. As you see, here we have no "commutator", but the best commutative quotient still exists.

Basically the same "commutatizing" construction works for all algebraic structures which have a binary operation in their type. For groups and rings, we are used to identify congruence relations with normal subgroups resp. ideals, and that's why $ab \sim ba$ gets "simplified" (really?) to $aba^{-1} b^{-1} \sim 1$ resp. $ab-ba \sim 0$.

• Nice categorical perspective, even if you don't categorical language. Do you have a source for communatizing with respect to category theory? I would like to read more on it. – PyRulez Jun 30 '15 at 3:04
• @MartinBrandenburg: Instructive answer! For the less trained: Could you also add a (short) paragraph and explain it in terms of category theory? +1 of course – Markus Scheuer Jun 30 '15 at 9:42
• Let me just say that "commutatizing" for some type of structure (containing at least one binary operation) is a functor which is left adjoint to the forgetful functor from the category of commutative structures to the category of structures. It always exists. – Martin Brandenburg Jul 1 '15 at 7:46
• Very, very informative. Thank you! – 6005 Sep 23 '15 at 17:24

Another perspective on this comes from considering the relationship between a Lie group and its corresponding Lie algebra. Details are at On the relationship between the commutators of a Lie group and its Lie algebra, but the upshot is that the ring commutator $[A,B]=AB-BA$ can be regarded as an infinitesimal version of the group commutator $[\alpha,\beta]=\alpha^{-1}\beta^{-1}\alpha\beta$.

In both cases, commutators measure how far the object is from being commutative.

A group is commutative when its operation is commutative.

A ring is commutative when its multiplication is commutative. (Addition in a ring is always commutative.)

Hence, the definition of commutator in each case reflects what it means to commute, as it should.

You can relate the commutator for a Lie group $G$ to the (ring-type) commutator of its Lie algebra $\mathfrak{g}$.

Say $g$ lies infinitesimally close to the group identity, $g = \exp(\epsilon\,X)$ for some $X \in \mathfrak{g}$. Likewise say $h = \exp(\epsilon\,Y)$ for some $Y \in \mathfrak{g}$. Then, $ghg^{-1}h^{-1}$ also lies infinitesimally near the identity of $G$. We compute using the Baker-Campbell-Hausdorff formula that its generator is $$\log ghg^{-1}h^{-1} = \epsilon^{2}[XY] + \mathrm{O}(\epsilon^3)\,.$$ If you want to make further contact with the ring commutator, think of the Lie bracket as "coming from" the commutator with respect to the associative product in $\mathrm{U}\mathfrak{g}$, the universal enveloping algebra of $\mathfrak{g}$, so that $$[XY] = XY - YX$$ in a genuine way.