Is this the way to solve the question?
Question a). Find the center of the group $S_3 \times \mathbb Z/6\mathbb Z$
Ans: $S_3$is the order of 6 element therefore, {1,(12),(23),(13),(123),(132)} and $\mathbb Z/6\mathbb Z$ is the integer with modulo of 6 (13 modulo 6 =1). The Center will be 1.
Question b). Show that for n > 2 the group $S_n$ is not cyclic, but can be generated by two elements.
Ans: 1) The group $S_n$ consists of all the bijective functions from the set {1, 2, ..., n} to itself. The group operation in $S_n$ is composition of functions. "Bijective" means one-to-one and onto. In particular, if F is an element of Sn and x and y are different elements of {1, 2, ..., n}, then F(x) and F(y) are not equal.
2) If n is at least 3, then $S_n$ contains the following two functions (and more):
The function A whose values are given by A(1) = 2, A(2) = 1, and A(x) = x when x is not 1 or 2.
The function B whose values are given by B(1) = 1, B(2) = 3, B(3) = 2, and B(x) = x for x not equal to 2 or 3.
3) We can compose these functions.
AB is the function whose values are given as follows: (AB)(1) = A(B(1)) = A(1) = 2 (AB)(2) = A(B(2)) = A(3) = 3 (AB)(3) = A(B(3)) = A(2) = 1 (AB)(x) = A(B(x)) = A(x) = x if x is not 1, 2, or 3 BA is the function whose values are given as (BA)(1) = B(A(1)) = B(2) = 3 (BA)(2) = B(A(2)) = B(1) = 1 (BA)(3) = B(A(3)) = B(3) = 2 (BA)(x) = B(A(x)) = B(x) = x if x is not 1, 2, or 3
4) Therefore it's not abelian, it cannot be cyclic.
Help me understand it better if I am wrong with an explanation.
http://en.wikipedia.org/wiki/Center_(group_theory)
The center of a group G, Z(G), is the set of all elements in G that are commutative with all elements in G. In a sense it is the "abelian part" of a group. If a group is abelian, then the center is the group itself, since the group is commutative. So there is more than just the identity in Z(GxH) in your question.