To prove that 1. implies 3., first note that any member of B is in the set (ker t + im q).
It is partially true: if a short exact sequence of groups is left split or a direct sum (1. or 3.
For a left split sequence, the map t × r: B → A × C gives an isomorphism, so B is a direct sum (3.
3. fails, because S3 is not abelian, but 2. holds: we may define u: C → B by mapping the generator to any two-cycle.
Note for completeness that 1. fails: any map t: B → A must map every two-cycle to the identity because the map has to be a group homomorphism, while the order of a two-cycle is 2 which can not be divided by the order of the elements in A other than the identity element, which is 3 as A is the alternating subgroup of S3, or namely the cyclic group of order 3.