Even permutations

Question

I am given the symmetric group $S_{9}.$

Let $$\sigma = \begin{bmatrix} 1 & 2& 3& 4& 5& 6& 7&8 &9 \\ 4& 8& 7& 9& 3& 1& 2& 5 & 6 \end{bmatrix}$$

Now the question asks:

Is $\sigma \in A_{9},$ the subgroup of even permutation on $1,2, \ldots,n$? Justify your answer.

Could someone fill me in as to what the question is asking? Sigma is a mapping so how is sigma an element in the set $A_9$?

$A_9$ is a subset of $S_9$ and the given $\sigma$ is an element of $S_9$. So it's either in $A_9$ or it's not. Which is it? That's the question. It's similar to a question such as: given that $4$ is an integer, is it even or is it not? — Casteels, Jun 12 '15 at 11:17
@Casteels Would you be kind to link me to the appropriate theorem (if it exists)? — Mathematicing, Jun 12 '15 at 11:21
The elements of $S_9$ and $A_9$ can indeed be realised as mappings from a nine element set to itself. It is these elements under composition which form the group. — Mark Bennet, Jun 12 '15 at 11:36
I think you are looking for the definition of an even permutation. This is a permutation which can be achieved by combining an even number of transpositions. A transposition is an element which exchanges two elements of the underlying set. Elements can be broken down into transpositions in many ways, but the decomposition will always have the same parity - so even and odd permutations are well-defined. Half the permutations are even, and half are off. The even permutations from a subgroup - the alternating group - of the full symmetric group. — Mark Bennet, Jun 12 '15 at 11:40
@MarkBennet I'm trying to link my understanding of the theorems to the question asked in the OP. The permutations in the above set in the OP is odd. How do I answer the question with this? — Mathematicing, Jun 12 '15 at 11:45
If you know it is an odd permutation then it isn't an even one, and you are done. You just have to show how you know that it is odd. — Mark Bennet, Jun 12 '15 at 11:54
Put it in another way, the question is asking if sigma is a subgroup I suppose? — Mathematicing, Jun 12 '15 at 11:57

score 0 · Accepted Answer · answered Jun 12 '15 at 11:22

0

Note that if we write $\sigma$ as a product of disjoint cycles we obtain $$\sigma = (1,4,9,6)(2,8,5,3,7)$$ i.e. an even and an odd permutation, and so $\sigma \not\in A_9$.

answered Jun 12 '15 at 11:22

WLOG

11,436

Correct me if I am wrong but the theorem I have on hand states that the permutation of a set is always either odd or even. I'll add that it is clear from the product of disjoint cycles that the disjoint cycles is an even number of 2-cycles. There are two 2-cycles. – Mathematicing Jun 12 '15 at 11:23
But if I were to express the above disjoint cycles in the form of products of transposition, we get seven 2-cycles. – Mathematicing Jun 12 '15 at 11:27
@Mathematicing Are you sure? You've likely made a mistake. – Casteels Jun 12 '15 at 11:29
Perhaps you could explain why the above is an odd and even permutation. According to Gallian (the book I'm using) it says that the permutation is always odd or always even. – Mathematicing Jun 12 '15 at 11:35
@Mathematicing: A cycle with an even number of entries is odd, and a cycle with an odd number of entries is even. You can search the proof – WLOG Jun 12 '15 at 11:42
I know that. I think my problem is answering the question with these theorem – Mathematicing Jun 12 '15 at 11:45
@Mathematicing: But the answer is in my post – WLOG Jun 12 '15 at 11:55

score 0 · Answer 2 · answered Jun 12 '15 at 11:26

One way that uses the representation of $\sigma$ that you are given is to look at the number of inversions: The total number of (ordered) pairs $(i,j)$ in the second row with $i>j$. If this number is even, then $\sigma\in A_9$. Otherwise $\sigma\not\in A_9$. In your case I count $21$ inversions: $(4,3)$, $(4,1)$, $(4,2)$, $(8,7)$, $(8,3)$, etc. So $\sigma\not\in A_9$.

Even permutations

2 Answers2