Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?
I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...
Sym(n) is the group that consists of all permutations of the first n natural numbers. sgn is the sign of the permutation.
I don't understand anything... S_n, A_n???
Has sym(n) even order?No, it isn't. If there exists such an onto homomorphism then the group has to have even order...and not all (finite) groups have even order, of course.
Tonio
I assume, and let NCA validated this, but since is surjective we have that and in particular that where is the number of cosets formed by (the index of in ) therefore since is the only subgroup of order it follows that for every surjection. But, this forces which shows that it is the only surjection.