So I have this question and i got pretty much stuck.

  1. Let $\pi$ be the permutation $$\pi= (1 2 3 4 5 6 7)\circ(1 3 5 7)\circ(2 4 6)$$ of the set $\{1,2,3,4,5,6,7\}$. Write $\pi$ as a product of disjoint cycles and determine if is an odd or even permutation.

I learnt how to do permutations(i guess) but from the whole permutation rows. Can someone help me uderstand what am i missing?

The solution is supposed to be $$\pi=(1 4 7 2 5)\circ(3 6)$$

Thanks a lot in advance!

  • $\begingroup$ It might help to rephrase this as "can you give a partition of $\{1,2,3,4,5,6,7\}$ such that $\pi$ preserves that partition?" $\endgroup$ – Couchy311 May 23 at 15:12
  • $\begingroup$ Welcome to MSE. You should learn some MathJax to format your questions. $\endgroup$ – saulspatz May 23 at 15:13

$1 \xrightarrow{\text{(246)}} 1 \xrightarrow{\text{(1357)}} 3 \xrightarrow{\text{(1234567)}}4 $

$4 \xrightarrow{\text{(246)}} 6 \xrightarrow{\text{(1357)}} 6 \xrightarrow{\text{(1234567)}}7 $

$7 \xrightarrow{\text{(246)}} 7 \xrightarrow{\text{(1357)}} 1 \xrightarrow{\text{(1234567)}}2 $

$2 \xrightarrow{\text{(246)}} 4 \xrightarrow{\text{(1357)}} 4 \xrightarrow{\text{(1234567)}}5 $

$5 \xrightarrow{\text{(246)}} 5 \xrightarrow{\text{(1357)}} 7 \xrightarrow{\text{(1234567)}}1 $

So we have $(14725)$

$3 \xrightarrow{\text{(246)}} 3 \xrightarrow{\text{(1357)}} 5 \xrightarrow{\text{(1234567)}}6 $

$6 \xrightarrow{\text{(246)}} 2 \xrightarrow{\text{(1357)}} 2 \xrightarrow{\text{(1234567)}}3 $

So we have $(36)$

Combining we have $(14725)\circ(36)$

$(14725)$ is even and $(36)$ is odd. The composition is odd.

  • $\begingroup$ Thanks for the great explanation, now i got how it works! $\endgroup$ – user10635779 May 24 at 17:41

The permutation $(2 4 6)$ is the function that maps $2$ to $4$, $4$ to $6$, $6$ to $2$, and leaves the other elements fixed. The permutations are composed from left to right. To see the effect of $\pi$ on $2$, for example, we see that the first permutation maps $2$ to $3$, the second permutation maps $3$ to $5$, and the last permutation maps $5$ to $5$ so that $\pi(2)=5.$ You need to do this for every element.

It is not a universal convention that the permutations compose from left to right. Indeed, I expected that they would compose from right to left, as functions generally do, but that gave the wrong answer.


Note that:

  • $\pi(1)=4$;
  • $\pi(4)=7$;
  • $\pi(7)=2$;
  • $\pi(2)=5$;
  • $\pi(5)=1$.

Now, you already have the cycle $(1\ \ 4\ \ 7\ \ 2\ \ 5)$. Among the elements of $\{1,2,3,4,5,6,7\}$, the ones that appear in it are $1$, $2$, $4$,$5$, and $7$. Now, start all over again, with an element that doesn't appear in it, such as $3$. You will get another cycle: $(3\ \ 6)$. And now there are no more elements left in $\{1,2,3,4,5,6,7\}$. So, $\pi=(1\ \ 4\ \ 7\ \ 2\ \ 5)\circ(3\ \ 6)$ indeed.

So, since $\pi$ is the composition of an even permutation with an odd one, it is an odd permutation.


You can get that $\pi=(14725)(36)$ just by checking where each element goes. That is, you have $1\to4\to7\to2\to5\to1$ and $3\to6\to3$.

Once you have that, it follows that $\pi$ is odd, since odd length cycles are even. For instance, $\pi=(15)(12)(17)(14)(36)$.

(It is a theorem that the parity of the number of transpositions a permutation can be written as the product of is an invariant. This invariant is called the sign of the permutation, $\bf{sgn}(\pi)$.)


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