What is the identity and inverse in a direct product of groups? 
If given the statement:
  $$ \text{G and H are groups such that }G \times H=\{{(x,y): x\in G  \text{ and } y\in H}\}$$
  How can I go about showing what the identity $e$ is as well as $a^{-1}$?

Here is my first attempt:
1) $$(x,y)*(e_1,e_2)=(x,y)$$
$$  (xe_1,ye_2)=(x,y)
                      $$ 
$$xe_1=x \implies e_1=1                        $$ 
$$                     \text{and}   $$ 
$$          ye_2=y \implies e_2=1              $$ 
$$                        (e_1,e_2)=(1,1))$$ 
from there I think I would need to generate the inverses from the following statements 
$$            (x,y)*(x^{-1},y^{-1})=(1,1)            $$ 
$$              (xx^{-1},yy^{-1})=(1,1)                                      $$$$ (xx^{-1})=1 \implies x^{-1} =\frac{1}{x}$$
$$       (yy^{-1})=1 \implies y^{-1} =\frac{1}{y}           $$                                  
Although this looks correct can anyone tell me if these look like all legal operations? 
 A: These answers aren't quite proving what you want.  For example, in the identity problem, you are supposing that an element $(a,b)$ has the property that 
$$
(x,y)*(a,b)=(x,y)
$$
and concluding that $a=1_G$ and $b=1_H$.  In other words, you've shown that if there exists an identity, then it must be of the form $(1_G,1_H)$.  Potentially, there are no elements of $G\times H$ that satisfy the equation above.  You haven't shown that $(1_G,1_H)$ is actually the identity.
Instead, you need to take your scratch work and turn it into a proof.  So, you need to prove that $(1_G,1_H)$ is the identity of $G\times H$ and that $(x,y)^{-1}=(x^{-1},y^{-1})$.  To prove the identity part, you must prove that for all $(x,y)\in G\times H$, 
$$
(x,y)*(1_G,1_H)=(x,y)
$$
and to prove the inverse part, you need to prove that for all $(x,y)\in G\times H$, 
$$
(x,y)*(x^{-1},y^{-1})=(1_G,1_H).
$$
You already have the proofs of these facts, but the assumptions/starting points are not right.
You don't need to show how you derived the identity and inverses, only that they enjoy the correct properties.
