What is the order of $\{0\} \times \mathbb Z_4$? What is the order of $\{0\} \times \mathbb Z_4$?
Not really sure what the order of this is?
$\{0\} \times \mathbb Z_4$ would be ?
$(0,0), (0,1), (0,2), (0,3)$? $4$?
 A: You've correctly listed the elements in $\{0\} \times \mathbb Z_4$, followed by the order of the group:
$$\{0\} \times \mathbb Z_4 = \{(0, 0), (0, 1), (0,2), (0,3)\} \implies |\{0\} \times \mathbb Z_4| = 4$$
Note that the order of the group also follows from the fact that $$|\{0\} \times \mathbb Z_4| = 1 \times 4 = |\{0\}| \times |\mathbb Z_4|$$
A: The order of a group is the number of elements it contains.  The cartesian product of finite groups $A,B$ has cardinality equal to the product of the cardinalities of $A,B$ respectively.  The trivial group $\{0\}$ has one element, while $\mathbb{Z}_4$ has four elements.  Consequently, the cartesian product $\{0\}\times \mathbb{Z}_4$ has four elements, and thus order 4.
A: It is not hard seeing that for groups $G$ and $H$ then $$e_H\times G=\{(e_H,g)\mid g\in G\}\cong G$$ For this, define $$\phi:~G\to H\times G\\ \phi(g)=(e_H,g)$$ and show that $\phi$ is a well-defined injective and $G\cong\phi(G)=e_H\times G$. Now lets doing some codes in GAP for your sample $\{0\}\times\mathbb Z_4$:
gap> Z4:=CyclicGroup(4);;
     StructureDescription(DirectProduct(Group(Identity(Z4)),Z4));


"C4"

