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In my homework, I was asked to determine the sign of the following permutation,

$\sigma =\begin{pmatrix} 1 &2 &3 &\cdots &n &n+1 &n+2 &\cdots &2n \\ 1 &3 &5 &\cdots &2n-1 &2 &4 &\cdots &2n \end{pmatrix}$

My plan: Firstly, I try this with some particular n to guess the rule for determining a certain sign related to the value n. Then, I prove it in a general way.

For $n=2$, we get: $\sigma =\begin{pmatrix} 1 &2 &3 &4 \\ 1 &3 &2 &4 \end{pmatrix}=\left ( 2\: 3 \right )\Rightarrow \text{sgn}(\sigma)=-1$.

For $n=3$, we get:$\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6\\ 1 &3 &5 &2 &4 &6 \end{pmatrix}=\left ( 2\: 3\:5\:4 \right )\Rightarrow \text{sgn}(\sigma)=-1$

For $n=4$, we get:$\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8\\ 1 &3 &5 &7 &2 &4 &6 &8 \end{pmatrix}=\left ( 2\: 3\:5\right )\left ( 4\: 7\:6\right )\Rightarrow \text{sgn}(\sigma)=1$

For $n=5$, we get:$\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 &10\\ 1 &3 &5 &7 &9 &2 &4 &6 &8 &10 \end{pmatrix}=\left ( 2\: 3\:5\:9\:8\:6\right )\left ( 4\:7\right )\Rightarrow \text{sgn}(\sigma)=1$

For $n=6$, we get: $\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12\\ 1 &3 &5 &7 &9 &11 &2 &4 &6 &8 &10 &12 \end{pmatrix}=\left ( 2\: 3\:5\:9\:6\:11\:10\:8\:4\:7\right )\Rightarrow \text{sgn}(\sigma)=-1$

For $n=7$, we get:$\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12 &13 &14\\ 1 &3 &5 &7 &9 &11 &13 &2 &4 &6 &8 &10 &12 &14 \end{pmatrix}=\left ( 2\: 3\:5\:9\:4\:7\:13\:12\:10\:6\:11\:8\right )\Rightarrow \text{sgn}(\sigma)=-1$

For $n=8$, we get:$\sigma =\begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12 &13 &14 &15 &16\\ 1 &3 &5 &7 &9 &11 &13 &15 &2 &4 &6 &8 &10 &12 &14 &16 \end{pmatrix}=\left ( 2\: 3\:5\:9\right )(\:4\:7\:13\:10)(6\:11)(8\:15\:14\:12)\Rightarrow \text{sgn}(\sigma)=1$

I get stuck here! It's not easy for me to understand how the rule works here! So, it's very nice to get your helpful suggestions. Please share with me if you have other ideas to solve this problem in easier way.

Many thanks

user26857
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Nguyễn Hữu Nam
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1 Answers1

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Hint: Instead of trying to compute the disjoint cycles representation, remember $$ \operatorname{sign}(\sigma)=\prod_{i<j}\frac{\sigma(i)-\sigma(j)}{i-j}=(-1)^{\#\{(i,j)\mid i<j\text{ and }\sigma(i)>\sigma(j)\}}. $$ and the set $$ \{(i,j)\mid i<j\text{ and }\sigma(i)>\sigma(j)\} $$ is very easy to count in this case.

user10354138
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  • Yes, use this definition is much easier to count! From your hint, I find out the number of inversions is $\frac{n(n-1)}{2}$, thus $\text{sgn}(\sigma)=(-1)^{\frac{n(n-1)}{2}}$. This result is fitted all cases of n I tried above! Thanks a lot! – Nguyễn Hữu Nam Oct 01 '23 at 04:50