Verified answer

Let b be a transposition (2-cycle) in Sn, and so Z2 = {1, b}.

Since An is normal in Sn (being of index 2), and An ∩ Z2 = {1} (since b is odd).

==> An * Z2 = Sn (clearly An * Z2 is a subgroup of Sn, but |An * Z2| = |Sn|, since An ∩ Z2 = {1}).

Therefore, Sn is a semidirect product of Z2 and the group of even permutations An.

I hope this helps!