# Proving that $x\ast y = f^{-1}(f(x)+f(y))$ defines a group operation.

I wasn't sure how to title this; it would either be too detailed or not detailed enough.

So I have an interval $I$ of $\mathbb R$ and $f : I \rightarrow \mathbb R$ is a bijection. Also, $\forall x,y \in I$, $x*y=f^{-1}(f(x) + f(y))$

I want to prove that $(I,*)$ is a group. My confusion is the fact that I don't know what the function actually is. For example, when finding the neutral element:

I want to find a $e\in I$ such that $\forall x\in I$, $x*e = e*x = x$

This means that I want to find a e such that $f^{-1}(f(x) + f(e)) = x$

Which means that I need to find an e such that $f(e) = 0$

but since I don't know what the function actually is, I can't do that.

I was also thinking that I could maybe prove that $f$ is an isomorphism, prove that $(\mathbb R, *)$ is a group, and thus so is $(I,*)$, but I don't think this is possible either. How should I be going about this problem? Thank you.

• You don't need to find numerically the element $e$ that evaluates to zero under $f$. What you need is to establish its existence, which in your case is guaranteed by the bijectiveness of $f$. Commented Apr 7, 2018 at 11:27
• $f$ is bijective, hence surjective, thus any real number is in its range, this gives us that $0$ is in its range, i.e. $\exists e$ s.t. $f(e)=0$.
– BAI
Commented Apr 7, 2018 at 11:30
• @BAI Oh, of course. Boy am I dumb. Thank you guys. Follow up question, how would I show associativity? Do I even have to? Commented Apr 7, 2018 at 11:37
• I changed the title. See if you like it. Commented Apr 7, 2018 at 11:43
• @user1551 I'm sorry if I'm being dense, but I don't think I know how to do this. I can say that $f(x') = -f(x)$, but I can't just apply the inverse function to both sides and get that $x' = f^{-1}(-f(x))$, right? Commented Apr 7, 2018 at 13:44

If you have an algebraic structure $A$ (semigroup, group, ring, field, etc), a set $X$, and a bijection $f\colon X\to A$, you can always use $f$ to define the same algebraic structure on $A$ by using $f$ to identify the elements of $X$ with elements of $A$ and operating there, and then using $f^{-1}$ to get the answer back to $X$.
In essence, it’s as if you have the structure written in Russian, $X$ is in English, and you use $f$ to translate the names of the elements of $X$ into Russian; clearly, if the operation forms a group/ring/field/module/monoid/lattice/whatever in Russian, then it will do so in English as well. Expliclty, if $\mu$ is an $n$-ary operation on $A$, you define the operation $\mu_X$ on $X$ by $$\mu_X(x_1,\ldots,x_n) = f^{-1}\Bigl(\mu\bigl(f(x_1),\ldots,f(x_n)\bigr)\Bigr).$$ Then $\mu_X$ will inherit all the required properties from $\mu$. All you are doing is using $f$ to “translate” from $X$ to $A$, where you already know things are “nice”.
Here, you have a group $\mathbb{R}$ under addition, and a bijection from your set $I$ to $\mathbb{R}$. You define an operation on $I$ by “take the elements to $\mathbb{R}$ using $f$, operate with them in $\mathbb{R}$, and then use $f$ to translate back the answer into $I$.”
For example, the fact that your operation is associative can be done as follows: given any $x,y,z\in I$, we know that $$f(x) + \bigl( f(y)+f(z)\bigr) = \bigl(f(x)+f(y)\bigr) + f(z).$$ To show that $x*(y*z) = (x*y)*z$, we have: \begin{align*} x*(y*z) &= f^{-1}(f(x)+f(y*z))\\ &= f^{-1}(f(x) + f(f^{-1}(f(y)+f(z))))\\ &= f^{-1}(f(x) + (f(y)+f(z)))\\ &= f^{-1}((f(x)+f(y))+f(z))\\ &= f^{-1}\Bigl( f(f^{-1}(f(x)+f(y))) + f(z)\Bigr)\\ &= (f^{-1}(f(x)+f(y))) * z\\ &= (x*y)*z. \end{align*} Similarly, the inverse of $x$ will be the $f^{-1}$-image of the inverse of $f(x)$; the neutral element will be the $f^{-1}$-image of the neutral element, etc.