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If $f$ is bijective and $g$ is bijective too, is $f(g(x))$ always same as $g(f(x))$.

My attempt:

I'm trying to look at an example if $f=x+1$ and $g=x+2$. The composition is commutative. But, i don't know in general.

If it's not, please show me the counterexample.

Thanks.

user516076
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2 Answers2

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Take $f(x)=x^3$ and $g(x)=x+1$

Then $g$ and $f$ are bijective functions from $\mathbb{R}\rightarrow \mathbb{R}$

But, $f(g(x))=(x+1)^3$ which is not the same as $g(f(x))=x^3+1$.

Akash Yadav
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If we consider $f(x)$, $g(x)$ bijective in a finite set of numbers then they can be considered as functions in a finite field such as $F_q$. Hence $f,g$ are permutations of the finite set of elements of $F_q$. Since $f(g(x))$ and $g(f(x))$ represent compositions of the two permutations, it is likely that permutations represented by $f,g$ do not commute. Since all functions in $F_q$ are polynomial functions all permutations of $F_q$ can be represented by polynomials (by interpolation). Hence to find such functions which do not commute, pick up any two permutations which do not commute and find their polynomial representations. This way you can get examples of non commuting polynomial compositions.

Viren Sule
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