Abstract Algebra: Parity of a Permutation

Abraham
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Homework Statement



How do I determine the parity of a permutation? I think my reasoning may be faulty.

By a theorem, an n-cycle is the product of (n-1) transpositions. For example, a 5 cycle can be written as 4 transpositions.

Now say I have a permutation written in cycle notation: (1 4 5)(2 3).

I say it is odd, because (1 4 5) can be written as two transpositions, and (2 3) is already a transposition, giving 3 total transpositions:

(1 4 5)(2 3) = (1 5)(1 4)(2 3).

Since the number of transpositions is odd, the permutation must be odd.
Agree or disagree?
 
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Abraham said:

Homework Statement



How do I determine the parity of a permutation? I think my reasoning may be faulty.

By a theorem, an n-cycle is the product of (n-1) transpositions. For example, a 5 cycle can be written as 4 transpositions.

Now say I have a permutation written in cycle notation: (1 4 5)(2 3).

I say it is odd, because (1 4 5) can be written as two transpositions, and (2 3) is already a transposition, giving 3 total transpositions:

(1 4 5)(2 3) = (1 5)(1 4)(2 3).

Since the number of transpositions is odd, the permutation must be odd.
Agree or disagree?

Agree. Note that you are implicitly using a very important and non-trivial theorem, which is that even though a permutation may be written in many different ways as a product of transpositions, the parity is the same no matter which product you choose.

For example, a 5-cycle may be written as a product of 4 transpositions, or 6 transpositions, or 8, etc., but there's no way to write it as a product of 5 or 7 or 9...
 
Thank you, for the swift reply.
 
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