# Index notation for group multiplication and commutators

Are there any serious errors in the following discussion of notation? I will be grateful for any corrections, suggestions and pointers towards the relevant literature.

The index notation for inner automorphisms: $$x^a \equiv_{\text{def}} a^{-1}xa$$ has proved a useful abbreviation, as also has the use of the index $^{-1}$ to denote the anti-involution taking an element to its inverse.

It seems sensible to allow: $$x^{-a} = (x^{-1})^a$$ Indeed, this abbreviation might assist newcomers to the study of groups to avoid the tempting confusion between $x^{-a}$ and $x^{a^{-1}}$.

We already have analogues of the multiplication laws for indices: $$(x^a)^b = x^{ab}$$ and $$x^ay^a = (xy)^a$$

Can we also introduce a further shorthand using a non-commutative addition?

The proposal (if it is not already in use) would be to set $$x^{a+b} \equiv_{\text{def}} x^ax^b$$ If $n$ is an integer it is interpreted as an ordinary power, so $$x^{1+a} \equiv x\cdot x^a$$ a commutator may be written \begin{align} (x,y) &= x^{-1}y^{-1}xy = y^{-x+1} = x^{-1+y} \end{align} with the corresponding \begin{align} (x,y)^{-1} &= y^{-1}x^{-1}yx = y^{-1+x} = x^{-y+1} \end{align} The triple commutator is a challenge even in linear notation, with $$(x,y,z) \equiv ((x,y),z) = y^{-1}x^{-1}yxz^{-1}x^{-1}y^{-1}xyz$$ Perhaps experimenting with index notation does not have much to offer as a practical tool, but it may assist beginners in learning to parse such expressions. The notation $(x,y,z)$ is admirably concise, but does not immediately conjure up a representation of the subtlety of the operations it encapsulates. $z$, the "last man in" sees the following: \begin{align} (x,y,z) & = (y^{-1}x^{-1}y)\cdot (xz^{-1}x^{-1}) \cdot (y^{-1}xy)z \\ &= x^{-y} z^{-x^{-1}} x^{y}z \\ &= z^{-x^{-1+y}+1} \end{align} From a different vantage point, viewing $y$ as the subject acted upon, we may see the triple commutator as \begin{align} (x,y,z) \equiv ((x,y),z) & = y^{-1}(x^{-1}yx)(z^{-1}x^{-1}y^{-1}xy z) \\ &= y^{-1+x+(-x+1)z} \\ &= x^{-y+1+(-1+y)z} \end{align} To conclude, here is an illustration of how even an ad hoc macro notation can sometimes be of assistance in unscrambling the rather long strings met with in commutator relations. It is a well-known (and rather remarkable) result that: $$(x,y^{-1},z)^y\cdot(y,z^{-1},x)^z\cdot(z,x^{-1},y)^x = 1$$ If we define, temporarily, $$\binom{a}{b,c} = aba^{-1}ca$$ then $$(x,y^{-1},z)^y = \binom{x^{-1}}{y^{-1},z^{-1}} \binom{y}{x,z} \\ (y,z^{-1},x)^z = \binom{y^{-1}}{z^{-1},x^{-1}} \binom{z}{y,x} \\ (z,z^{-1},y)^x = \binom{z^{-1}}{x^{-1},y^{-1}} \binom{x}{z,y}$$ The result now follows, since, as we see from the definition $$\binom{a}{b,c}^{-1} = \binom{a^{-1}}{c^{-1},b^{-1}}$$