1
$\begingroup$

Hi I'm having trouble with this homework question: let $\eta$ (below) belong to $S_n$

$$ \begin{matrix} 1 & 2 & ... & n-1 & n \\ n & n-1 & ... & 2 & 1 \\ \end{matrix} $$

Determine the sign of $\eta$ given that $n$ is even

So i know that to calculate the sign of $S_n$ i need to use the formula:$(−1)^{(s1−1)+(s2−1)+...}$ where $s1, s2... $ are the lengths of the cycles. However, when I use the formula I get different answers i.e. I get $-1$ for $n=2$, $1$ for $n=4$. Am i right in thinking that if (for $n=2$) then its just $<1,2>$ so it's $(-1)^1$ $=-1$. Then for $n=4$ it's $<1,4>$ $<2,3>$ then it's $(-1)^{2-1+2-1}$ $=1$.

Am i using the wrong numbers for $s1$ and $s2$?

Many thanks

  • 1
    If you write the permutations in cycle notation, even cycles are odd, odd cycles are even and normal rules of multiplication apply (for example $(12)(34)$ is odd times odd, so it's even) – 2017-02-15
  • 1
    You've observed (for small $n$, but it's true in general) that $\eta$ is a product $(1, n) (2, n - 1) \cdots \left(\frac{n}{2}, \frac{n}{2} +1\right)$ is a product of $\frac{n}{2}$ $2$-cycles (transpositions), so $s_1 = s_2 = \cdots s_{n / 2} = 1$, just as you claim. – 2017-02-15
  • 1
    Hint. Can you see how to write this permutation as a product of transpositions? – 2017-02-15

1 Answers 1

1

We can do it for any $n$ in general. Notice that there are $\lfloor n/2\rfloor$ orbits of size $2$, and possibly one fixed element. Therefore the sign of the permutation is $-1^{\lfloor n/2 \rfloor}$