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What is the sign of the following permutation. Prove your answer:

$$\pmatrix{1 & 2 & \cdots&p&p+1&\cdots & \cdots &p+q \\ q+1 & \cdots & \cdots & q + p& 1 & \cdots& \cdots &q}.$$

I said let $\tau$ equal the permuation. Then, we get sign$(\tau) = (-1)^{pq}$. We can show this by noticing that $\tau$ shifts $a \mapsto a + q$. This then gives us that $\tau = \pmatrix{1 & 2 & \cdots (p+q)}^q$. S0 from here we get

$$(-1)^{\tau} = \left((-1)^{p+q-1}\right)^q= \left((-1)^{pq + (q-1)q}\right).$$

$(q-1)q = \mathrm{odd} \, \times \, \mathrm{even} = \mathrm{even}$, which doesn't influcence the power much and so we get

$$(-1)^{pq + (q-1)q}\ = (-1)^{pq}$$

which proves my answer.

Is this correct?

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Kaish
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  • Presumably there is a typo, and $p$ is supposed to map to $p + q$. – Ben Millwood Apr 19 '13 at 19:59
  • Now that makes sense! – amWhy Apr 19 '13 at 20:00
  • I also think the parentheses in $\tau = \pmatrix{1 & 2 & \cdots (p+q))^q}$ are off. I think you want $\tau = \pmatrix{1 & 2 & \cdots (p+q)}^q$ – Ben Millwood Apr 19 '13 at 20:01
  • @BenMillwood Oh yeah, I see the typo. Edited. Oh, no, I thought the $q$ went outside the brackets. Why would it go inside? – Kaish Apr 19 '13 at 20:02
  • Kaish, it looks like there are mismatched parentheses: an additional unmatched parenthesis on the right side of your expression for $\tau$. The exponent $q$ does go outside the cycle: Ben was noticing the extra parenthesis following the exponent. – amWhy Apr 19 '13 at 20:11
  • Your argument is correct. You can also observe that the number of pairs of integers, $i<j$, whose order is reversed (i.e. $\tau(i)>\tau(j)$) is clearly $pq$, because reversal happens, iff $i\le p$ and $j>p$: $p$ choices for $i$ and $q$ choices for $j$. – Jyrki Lahtonen Apr 19 '13 at 20:55

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You are correct: $$\tau = \pmatrix{1 & 2 & \cdots (p+q)}^q,\;\text{and} \;\operatorname{sign}(\tau) = (-1)^{pq}.$$

Alternatively, we can notice that there are exactly $pq$ integer pairs $$(m, n),\; m < n,\; \; m, n \in \{1, 2, \cdots , (p+ q)\}\;$$ which are reversed (switched) under $\tau$ so that $\tau(m) > \tau(n),\;$ where $\;m \leq p$ and $n>p$.

That gives us $p$ choices for $m$ and similarly, $q$ choices for $n$: which gives a total of $\,pq\,$ such reversals/switches under $\tau$, and hence $$\;\operatorname{sign}(\tau) = (-1)^{\large pq}$$

amWhy
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