Even permutation vs even function

Question

We know that a permutation $\sigma$ is a bijection on $\{1,2,\cdots, n\}$. Also a function $\sigma$ is even if $\sigma(-x)=\sigma(x)$ for all $x$. I'm wondering is there a relation between even permutation and even function?

Answer 1

As @AndresMejia points out, it's not immediately obvious what $-x$ means. There are natural suggestions, like $-x := n - x$, ie the additive inverse modulo $n$.

However, as you note, if $\sigma : \{1, ..., n\} \to \{1, ..., n\}$ is a permutation, then it is a bijection (that is the definition). So whatever $-x$ means, we must have $x = -x$ if $\sigma(x) = \sigma(-x)$. (To see this, note that the inverse $\sigma^{-1}$ exists, as $\sigma$ is a bijection on a finite set. Then $\sigma(x) = \sigma(-x)$ implies that $x = (\sigma^{-1} \circ \sigma)(x) = (\sigma^{-1} \circ \sigma)(-x) = -x$.)

So basically unless you are in a situation where $x = -x$ for all $x$, then a function can't be both bijective and even (at least on a finite set).

Answer 2

There is only a distant relationship in that "the even elements form a cone," by which I mean the product of two even things is again even.

In the case of even functions, the product is the pointwise product: $(f\cdot g)(x):=f(x)\cdot g(x)$, and in the case of permutations it is composition of permutations.

For odd functions and permutations, we also have the commonality: the product of odds is even.

Finally, the product of an even and an odd with be odd.

You'll notice there are also differences, though, too. At the very least, not every function is even or odd, but every permutation on a finite set is even or odd. Secondly are the operations I mentioned earlier: the function multiplication of real-valued functions is always commutative, but permutation composition is quite often noncommutative. Permutations form a group, but the functions (under multiplication) do not.