Let a,b be in S_n.

Show a,ba^-1b^-1 is an element of A_n

aba^-1b^-1

a(ba^-1)b^-1

a(ab)b^-1

aa(bb^-1)

aa

=e

Have no clue if what I did was correct

A_n is alternating group

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- Nov 8th 2010, 05:34 PMkathrynmathShow a,ba^-1b^-1 is an element of A_n
Let a,b be in S_n.

Show a,ba^-1b^-1 is an element of A_n

aba^-1b^-1

a(ba^-1)b^-1

a(ab)b^-1

aa(bb^-1)

aa

=e

Have no clue if what I did was correct

A_n is alternating group - Nov 8th 2010, 06:15 PMtonio
- Nov 8th 2010, 06:42 PMkathrynmath
even times even=even

odd times odd = odd - Nov 8th 2010, 06:46 PMtonio
- Nov 8th 2010, 06:52 PMkathrynmath
Ok a permutation and it's inverse have the same parity.

aba^-1b^-1

So a*a^-1=even

and b^-1*b=even

or

a*a^-1=even

and b^-1*b=odd

or

a*a^-1=odd

and b^-1*b=even

or

a*a^-1=odd

and b^-1*b=odd - Nov 8th 2010, 07:10 PMkathrynmath
I don't know if that was the right idea

- Nov 8th 2010, 07:35 PMtonio
- Nov 8th 2010, 07:55 PMkathrynmath
Oh, I hadn't thought of writing it that way. Since they are the same parity, they must be in A_n?

- Nov 8th 2010, 09:21 PMtopspin1617
No no no, not for permutations. The rules for parity for multiplying permutations are analagous to the rules for ADDING integers.

- Nov 9th 2010, 01:42 AMtonio