Proving isomorphism between two sets of ordered pairs. Suppose I have a set of ordered pairs whose first elements are members of $(G,*)$ and whose second elements are members of $(H,\#)$, where $G,H$ are arbitrary sets and $*,\#$ arbitrary binary operations.  I have to show that $G\times H \cong H\times G$.
So my assumption is that I have to show thre exists an isomorphism betwixt the two groups.  So I define $$\varphi:G\times H\rightarrow H\times G$$ $$\varphi(g,h)=(h,g)$$
Now I have to show both injectivity and surjectivity.  Injectivity by
$$\varphi(g_1,h_1)=\varphi(g_2,h_2)$$
$$(h_1,g_1)=(h_2,g_2)$$
The fact that these two ordered pairs have elements that are members of groups i think needs to be utilized, but I'm not seeing how...any hints?
 A: You have shown that $\varphi$ is a bijection, which is a good start.  Just as a homomorphism is a function which preserves structure, an isomorphism is a bijection which preserves and reflects structure.  (This is an oversimplification in general, but is accurate for these sorts of algebraic structures). 
So what structure is there?  The two binary operations!  So, we have a binary operation $\cdot$ on $G\times H$ defined by:
$$\langle g_1,h_1\rangle \cdot \langle g_2,h_2\rangle=\langle g_1\ast g_2,h_1\# h_2\rangle$$
and a corresponding operation $\square$ on $H\times G$. 
To show that $\varphi$ is an isomorphism, you have to show that 
$$\varphi(\langle g_1,h_1\rangle\cdot \langle g_2,h_2\rangle)=\varphi(\langle h_1,g_1\rangle)\square \varphi(\langle h_2,g_2\rangle)$$
A: Another way to see bijectivity is note that $\varphi \circ \varphi$ is the identity map, so $\varphi$ is its own inverse.  A function is bijective if and only if it has an inverse.
To show that you have an isomorphism, you have to show that the multiplication is preserved by $\varphi$.  Let us denote the multiplication on $G \times H$ by $\otimes$, i.e $(g_1, h_1) \otimes (g_2, h_2) = (g_1 * g_2, h_1 \# h_2)$.  Similarly, let $\otimes'$ denote multiplication in $H \times G$.  Then you must show that
$$
\varphi((g_1, h_1) \otimes (g_2, h_2)) = \varphi(g_1,h_1) \otimes' \varphi(g_2,h_2).
$$
This turns out to be a straightforward unpacking of all of the definitions given.
