Do odd permutations of S_n form a group? Explain?
I know that they DO NOT because the identity is even. But I dont know how to show that.
2007-09-27 03:22:32 · 4 answers · asked by ClooneyIsAGenius 2 in Science & Mathematics ➔ Mathematics
I'm sure because we talked about it in class, but I'm trying to figure out how to SHOW odd perms don't contain the identity.
2007-09-27 03:29:42 · update #1
It IS enough.
I'm saying this... I know that it's not a group for the reason that there is no identity! BUT i want to show why there is no identity in the set.
2007-09-27 03:35:34 · update #2
Every permutation can be written as a product of transpositions (cycles of length 2) -- obviously not in a unique way. Even if this decomposition is not unique, it's parity is well defined - i.e. you can write it as a product with either an even or an odd number of permutation.
A permutation is even iff you can write it as the product of an even number of transpositions.
The identity is even because you can write it e.g. as (1 2)(1 2), and since it's even it can't be odd.
Please also note that the product of two odd permutations is even, so you can't have a subgroup.
You can find everything you need about this subject basically on every Algebra book.
2007-09-27 03:58:02 · answer #1 · answered by eflags9x86 1 · 1⤊ 0⤋
Isn't that enough? The subset of odd permutations does not include the identity. All groups must have an identity element. Therefore, the subset of odd permutations cannot be a group.
2007-09-27 03:30:38 · answer #2 · answered by acafrao341 5 · 1⤊ 0⤋
eflags is right
2007-09-27 12:14:20 · answer #3 · answered by Theta40 7 · 0⤊ 0⤋
If you can't show it, how can you be sure?
2007-09-27 03:27:31 · answer #4 · answered by Anonymous · 0⤊ 2⤋