What is the identity and inverse in a direct product of groups?

Question

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?

Answer 1

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.