Group Isomorphic problem 
Prove that for any groups, $G$ and $H$, we have $$(G \times H)/G' ≅ H\tag{1}$$ and $$(G \times H)/H' ≅ G,\tag{2}$$ where $G' =\{(g,e) \in G \times H | g \in G\}$ and $H' =\{(e,h) \in G \times H | h \in H\}$.

I tried to first expand this by plugging the identities in but I couldn't find anything to go off of. Is there some simple thing I'm missing?
Thank you in advance!
 A: First make sure you understand why $G'$ and $H'$ are normal subgroups.
We will demonstrate a very simple example of the first isomorphism theorem. Namely:
Define $\pi: G \times H \to H$ by $\pi(g,h) = h$. It is clear that $\pi$ is surjective, and make sure you understand why it is a homomorphism of groups.
Now what is its kernel? It is the set $\{(g,h) \in G \times H : h = e\}$ which is $G'$.
The first isomorphism theorem then gives $(G \times H) / G' \cong H$ (and the other claim is symmetric, now).
Now in case you haven't yet learned the isomorphism theorems we can do:


*

*Show $G'$ is normal in the product $G \times H$. Hence $(G \times H) / G'$ has the structure of a group.

*Note that $(g_1, h_1)G' = (g_2, h_2)G'$ if and only if $(g_1 g_2^{-1}, h_1 h_2^{-1}) \in G'$, or $h_1 = h_2$, simply.
This implies we can define a group homomorphism:


*Define $\psi:(G \times H) / G' \to H$ by $\psi((g,h)G') = h$. To make the point clear, this assignment is well defined because $(g_1,h_1)G' = (g_2,h_2)G'$ if and only if $h_1 = h_2$. Now show this function is a homomorphism, surjective and injective. Let's show injectivity, for example:


$h = e_H$ implies that for any $g \in G$, $(g,h)G' = (g, e_H)G' = G'$.
Hope this clarifies.
