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

• Nov 8th 2010, 05:34 PM
kathrynmath
Show 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 PM
tonio
Quote:

Originally Posted by kathrynmath
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

Even permutation times even permutation = ...? Odd permutation times odd permutation = ?

Tonio

Ps. And no, what you did isn't correct. There's no commutativity in S_n , for n >= 3
• Nov 8th 2010, 06:42 PM
kathrynmath
even times even=even
odd times odd = odd
• Nov 8th 2010, 06:46 PM
tonio
Quote:

Originally Posted by kathrynmath
even times even=even
odd times odd = odd

Good, so now just remember that a permutation and its inverse have the same parity...and now solve your problem!

Tonio
• Nov 8th 2010, 06:52 PM
kathrynmath
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 PM
kathrynmath
I don't know if that was the right idea
• Nov 8th 2010, 07:35 PM
tonio
Quote:

Originally Posted by kathrynmath
I don't know if that was the right idea

**** Sigh **** (Crying).... $[a,b]=aba^{-1}b^{-1} = (ab)(ba)^{-1}$ ... and thus we only care whether ab (and ALSO ba) is even or odd...

Tonio
• Nov 8th 2010, 07:55 PM
kathrynmath
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 PM
topspin1617
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 AM
tonio
Quote:

Originally Posted by kathrynmath
Oh, I hadn't thought of writing it that way. Since they are the same parity, they must be in A_n?

No! No matter what parity $ab, (ba)^{-1}$ have, when you multiply them the parity is even and thus the product is in $A_n$...(Angry)(Headbang)

Tonio