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
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
Last edited by tonio; Nov 8th 2010 at 06:16 PM.
Reason: Addition
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
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 $\displaystyle ab, (ba)^{-1}$ have, when you multiply them the parity is even and thus the product is in $\displaystyle A_n$...