2

In a homework assignment, I am asked to find $(P_3 \circ P_1)^{-1}$ knowing:

Let $P_1 = (3\ 4\ 1\ 2\ 5), P_2 = (3\ 5\ 1\ 2\ 4)$ and $P_3 = (5\ 1\ 4\ 2\ 3)$ be three permutations.

I am second-guesing the final guessing the final step (where the composition is inverted), since the inversion seems to be the same value as the non-inverted set.

Is my method correct?

$(P_3 \circ P_1)^{-1}$ can be written as $(P_3(P_1(x)))^{-1}$ for each $x\in P_3$. So:

$(P_3 \circ P_1(1)) = (P_3(P_1(1))) = (P_3(3)) = 4$

$(P_3 \circ P_1(2)) = (P_3(P_1(2))) = (P_3(4)) = 2$

$(P_3 \circ P_1(3)) = (P_3(P_1(3))) = (P_3(1)) = 5$

$(P_3 \circ P_1(4)) = (P_3(P_1(4))) = (P_3(2)) = 1$

$(P_3 \circ P_1(5)) = (P_3(P_1(5))) = (P_3(5)) = 3$

So, $(P_3 \circ P_1) = (4\ 2\ 5\ 1\ 3)$ and so $(P_3 \circ P_1)^{-1} = (4\ 2\ 5\ 1\ 3)$

Lord_Farin
  • 17,743
Tim
  • 219

2 Answers2

2

I'm assuming that we are using cyclic notation: if $P=(a_1\ a_2\cdots a_n)$ ($a_i$ distinct), then $P$ is the permutation such that $P(a_i)=a_{i+1}$, $P(a_n)=a_1$, and $P(x)=x$ for $x\neq a_i$.

Given a permutation $P=(a_1\ a_2\cdots a_n)$, its inverse is given by simply inverting the order of the $a_i$'s, that is, $P^{-1}=(a_n\cdots a_2\ a_1)$. Here's how I'd solve the problem:

If we want to find $P_3P_1$, we start with $1$: Notice that $P_3P_1(1)=P_3(2)=3$, $P_3P_1(3)=P_3(4)=2$, and $P_3P_1(2)=1$, so $P_3P_1$ is the cycle $(1\ 3\ 2)$ possibly multiplied by something. Since the only elements left are $4$ and $5$, and $P_3P_1(4)=P_3(1)=4$, and $P_3P_1(5)=P_3(3)=5$, then $P_3P_1=(1\ 3\ 2)$. Therefore, $$(P_3\circ P_1)^{-1}=(2\ 3\ 1)=(1\ 2\ 3)$$

Luiz Cordeiro
  • 18,513
1

You are right to second-guess the last line. $(P_3\circ P_1)^{-1}$ should have the cycle of $(P_3\circ P_1)$, yet in reverse. That is: $$(P_3\circ P_1)^{-1} = (3\;1\;5\;2\;4)$$

More examples of finding the inverse of a cycle can be found here.

Community
  • 1
apnorton
  • 17,706