Etydhv

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

3. 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!

$1 xrightarrowtext(246) 1 xrightarrowtext(1357) 3 xrightarrowtext(1234567)4 $

$4 xrightarrowtext(246) 6 xrightarrowtext(1357) 6 xrightarrowtext(1234567)7 $

$7 xrightarrowtext(246) 7 xrightarrowtext(1357) 1 xrightarrowtext(1234567)2 $

$2 xrightarrowtext(246) 4 xrightarrowtext(1357) 4 xrightarrowtext(1234567)5 $

$5 xrightarrowtext(246) 5 xrightarrowtext(1357) 7 xrightarrowtext(1234567)1 $

So we have $(14725)$

$3 xrightarrowtext(246) 3 xrightarrowtext(1357) 5 xrightarrowtext(1234567)6 $

$6 xrightarrowtext(246) 2 xrightarrowtext(1357) 2 xrightarrowtext(1234567)3 $

So we have $(36)$

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

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

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 $1to4to7to2to5to1$ and $3to6to3$.

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, $bfsgn(pi)$.)

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.