# Question in discrete mathematics about group permutations

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 $$\pi=(1 4 7 2 5)\circ(3 6)$$

• 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?" – Couchy311 May 23 at 15:12
• Welcome to MSE. You should learn some MathJax to format your questions. – 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.

• Thanks for the great explanation, now i got how it works! – 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, $$\bf{sgn}(\pi)$$.)