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In this case, $x$ and $f(x)$ belong to a multiplicative group. I would like to write $x f(x)^{-1}$ instead of $x(f(x))^{-1}$. I figured that I could explain once what I mean by $x f(x)^{-1}$ and then use it freely after that. Is this acceptable?

Steven Alexis Gregory
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    By $f(x)^{-1}$, do you mean $f^{-1}(x)$ or $1\over f(x)$? – SS_C4 Mar 14 '16 at 08:04
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    Introducing notation and then using it consequently in a correct way should be ok. Just avoid ambiguity to already existing notation. – Imago Mar 14 '16 at 08:05
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    $xf(x)^{-1}$ (or $x,f(x)^{-1}$ with a small space) seems fine to me as long as $x$ is an element of the group and $f$ is a function. $x(f(x))^{-1}$ looks superfluous a bit. Another option would be $x\cdot f(x)^{-1}$ where $\cdot$ denotes the multiplication. – Blackbird Mar 14 '16 at 08:54

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I had exactly the same problem recently and I decided to opt for brackets, as in the expression $$ \Delta^{abb} = [f(bb)]^{-1} f(b) [f(1)]^{-1} f(b) [f(ab)]^{-1} f(a) [f(ab)]^{-1} f(abb) $$ An alternative solution is to have an additive notation for the group (even if it is not commutative, but better warn the reader), which would give $$ \Delta^{abb} = - f(bb) + f(b)-f(1)+f(b)-f(ab)+ f(a)-f(ab) + f(abb) $$

J.-E. Pin
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  • In my case, I wanted to emphasize that the general theory worked for all groups and then I wanted to handle the special case of finite abelian groups using additive notation. Judging from the other comments about my question, as much as I wish I didn't have to do it, It looks like brackets is the best answer. – Steven Alexis Gregory Mar 14 '16 at 17:04
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An alternative ist just using maps without argument variables, like $$ \DeclareMathOperator{id}{id} \id f^{-1} $$ if you mean the product, or $$ \id \circ f^{-1} $$ if you mean the composition of $\id$ and the inverse of $f$. Or did you mean $$ \id / f $$

mvw
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