Question in discrete mathematics about group permutations

discrete mathematicsgroup-theorypermutation-cyclespermutations

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!

Best Answer

$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.

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