Find an example of an onto group homomorphism from Sn to Z2, and determine its kernel
Solution
There are only two group homomorphisms from Sn to Z2 for any n>1. The trivial
mapping, which sends every element of Sn to 0, is one of them. The other sends every even
permutation to 0 and every odd permutation to 1. To see why, recall the first isomorphism
theorem. If f is some homomorphism from Sn to Z2, then Sn/ker(f) is isomorphic to f(Sn). If we
assume that f isn\'t the trivial mapping, then there is some element of Sn which maps to 1, which
means f(Sn) = Z2. Furthermore, if f is a non-trivial homomorphism, then every element of Sn
that maps to 1 under f is contained in the coset xker(f), where x is some element such that f(x) =
1. Thus, the number of elements that map to 0 is the same as the number of elements that map to
1. Since the kernel of a homomorphism is always a normal subgroup, this means we are looking
for all normal subgroups of Sn whose order is exactly half of |Sn|. However, for each Sn (where
n>1), the only such normal subgroup is An, the set of all even permutations in Sn. Thus, for any
non-trivial homomorphism f, all even permutations map to 0 and all odd permutations map to 1..
Find an example of an onto group homomorphism from Sn to Z2, and det.pdf
1. Find an example of an onto group homomorphism from Sn to Z2, and determine its kernel
Solution
There are only two group homomorphisms from Sn to Z2 for any n>1. The trivial
mapping, which sends every element of Sn to 0, is one of them. The other sends every even
permutation to 0 and every odd permutation to 1. To see why, recall the first isomorphism
theorem. If f is some homomorphism from Sn to Z2, then Sn/ker(f) is isomorphic to f(Sn). If we
assume that f isn't the trivial mapping, then there is some element of Sn which maps to 1, which
means f(Sn) = Z2. Furthermore, if f is a non-trivial homomorphism, then every element of Sn
that maps to 1 under f is contained in the coset xker(f), where x is some element such that f(x) =
1. Thus, the number of elements that map to 0 is the same as the number of elements that map to
1. Since the kernel of a homomorphism is always a normal subgroup, this means we are looking
for all normal subgroups of Sn whose order is exactly half of |Sn|. However, for each Sn (where
n>1), the only such normal subgroup is An, the set of all even permutations in Sn. Thus, for any
non-trivial homomorphism f, all even permutations map to 0 and all odd permutations map to 1.
