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$\alpha _n ^n-1=0$

$\alpha _n=e^{2 \pi i/n}$

$$f(x_1,x_2,x_3,\ldots,x_n)=(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n$$ Prove that $f(x_1,x_2,x_3,\ldots,x_n)$ is not symmetric function for $n>4$

for example $n=2$

$\alpha _2=e^{2 \pi i/2}=-1$

$$f(x_1,x_2)=(x_1+\alpha _2 x_2)^2=(x_1-x_2)^2=x_1 ^2 + x_2 ^2 -2 x_1 \cdot x_2$$

$$f(x_1,x_2)=f(x_2,x_1)=x_1 ^2 + x_2 ^2 -2 x_1 \cdot x_2$$

Therefore, for $n=2$, $f$ is a symmetric function

I noticed such functions while working on the proof of Abel–Ruffini theorem. The theorem shows that there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher . In the book, I also noticed symmetric functions definitions. This function has a key role for showing that there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher.

I would like to learn how to prove that $n>4$ the $f(x_1,x_2,x_3,\ldots,x_n)$ is not symmetric function without using group theory because I am new learner in the subject. If possible please show in similiar way as I shew for $n=2$.

Thanks for answers and your time.

EDIT: After asking the question, I have noticed one property for the function. I wanted to share it.

$f(x_1,x_2,x_3,\ldots,x_n)= (x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n= $ $$=\alpha _n ^{n}(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n=(x_n+\alpha _n x_1+ \alpha _n ^2 x_2+\cdots+\alpha _n ^{n-1} x_{n-1})^n=\alpha _n ^{n}(x_n+\alpha _n x_1+ \alpha _n ^2 x_2+\cdots+\alpha _n ^{n-1} x_{n-1})^n=(x_{n-1}+\alpha _n x_n+ \alpha _n ^2 x_1+\cdots+\alpha _n ^{n-2} x_{n-2})^n=....$$ If do this steps n times we will get same function in the begining, Thus

(totally $n$ permutation of f is equal each other)

but Total permutation of $(x_1,x_2,x_3,\ldots,x_n)$ is $n!$

I can see that only for n=2 the permutations are equal. So I need to extend the statement for n>2 .. And also $Generic Human$ got same result with different approach . Thanks a lot for your time.

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I think you can prove it by showing that $f(x_1,x_2,0,0,...,0)\neq f(x_2,x_1,0,0,...,0)$ using much the same proof. – Thomas Andrews Mar 13 '12 at 20:41
up vote 4 down vote accepted

Suppose $f$ is symmetric. \begin{aligned} f(1,x,0,...,0)&= (1+\alpha_n x)^n \\ =f(x,1,0,...,0)&= (x+\alpha_n)^n \\ &= (1+\alpha_n^{-1} x)^n \end{aligned}

This is a smooth function, so we can compute the derivative with respect to $x$ at $x=0$: $$n\alpha_n = n\alpha_n^{-1}$$ This implies that either $n=0$ or $\alpha_n^2 = 1$, that is $n \leq 2$. So the theorem is actually true for all $n > 2$.

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