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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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
Quote:
Originally Posted by kathrynmath
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
even times even=even
odd times odd = odd
Quote:
Originally Posted by kathrynmath
Good, so now just remember that a permutation and its inverse have the same parity...and now solve your problem!
Tonio
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
I don't know if that was the right idea
Quote:
Originally Posted by kathrynmath
**** Sigh **** (Crying)....... and thus we only care whether ab (and ALSO ba) is even or odd...
Tonio
Oh, I hadn't thought of writing it that way. Since they are the same parity, they must be in A_n?
No no no, not for permutations. The rules for parity for multiplying permutations are analagous to the rules for ADDING integers.
Quote:
Originally Posted by kathrynmath
No! No matter what parityhave, when you multiply them the parity is even and thus the product is in
...(Angry)(Headbang)
Tonio