The kernel of a group homomorphism is
The homomorphism sends identity to identity and inverses to inverses. Thus , and if , then , so is a subgroup by the subgroup test.
For any and ,
Hence . Applying the same inclusion with gives equality, proving
This normality argument does not require either group to be finite.
The conjugacy class of is
If is a normal subgroup and , then normality gives for every . Thus the whole conjugacy class of each member of is contained in , and
Conjugacy classes partition the group, so this is a union of complete disjoint classes, never selected fractions of classes.
The cycle type records the lengths of the disjoint cycles in a permutation, including fixed points as length-one cycles. The possibilities for three objects are
These exhaust the six elements of the symmetric group , denoted in the source. They are also the conjugacy classes of the symmetric group on three letters.
The kernel is a normal subgroup, so it contains the identity class and is a union of the classes listed in part (c). Its candidate orders are therefore , , and . Lagrange's theorem excludes four, since it does not divide six. The remaining possibilities are
All three are normal: the first and last trivially, while is the kernel of the map giving the sign of a permutation, or the index-two subgroup generated by a 3-cycle.
Because is onto, the first isomorphism theorem gives . Explicitly, the map is well defined, multiplicative and surjective; equality of its values means , so it is also injective. Thus
Each is realized by the identity map, the sign map and the trivial map, respectively. This completes the classification of normal subgroups and quotients of the symmetric group on three letters.

Articles by others on the same topic (0)

There are currently no matching articles.