Prove that all groups with three elements are isomorphic The Cayley table clearly states that there is only one possible "blueprint" of a group with three elements. However, in order to prove that groups are isomorphic, we need to prove that they are homomorphic, and so if we take a function $f:G_1\rightarrow G_2$,
$$f(x *_1y) = f(x)*_2f(y)$$
What kind of operations should we define for the three-element groups to prove that they are in fact homomorphic?
 A: If you want to show that two arbitrary three-element groups are isomorphic, then you cannot choose what operations they have. However that doesn't mean that you can't "discover" their operations. Suppose $G:=\{e,x,y\}$ is a three-element group. What are the possibilities for $xy$?
If $xy=x$, then $y=e$ is a contradiction. Similarly $xy=y$ gives the contradiction $x=e$. Thus $xy=e$, which also gives $yx=e$.
Let's do the same thing for $x^2$:
If $x^2=e$, then $x=x^{-1}$ is a contradiction to $x^{-1}=y$ obtained above. If $x^2=x$, then $x=e$ is a contradiction. Thus $x^2=y$.
Similarly we can obtain that $y^2=x$. Therefore we know exactly how the operation on $G$ behaves. Let's use this to define an isomorphism.
Let $G_1:=\{e_1,x_1,y_1\}$ and $G_2:=\{e_2,x_2,y_2\}$ be two three-element groups. Define $f:G_1\to G_2$ by $f(e_1)=e_2$, $f(x_1)=x_2$, and $f(y_1)=y_2$.
Then $f$ is clearly a bijection. To see that it is a homomorphism, we must verify
$$
f(x_1y_1)=f(x_1)f(y_1), \quad
f(y_1x_1)=f(y_1)f(x_1),\quad f(x_1^2)=f(x_1)^2,\quad\text{and}\quad f(y_1^2)=f(y_1)^2,
$$
as well as the (more obvious) identities involving $e_1$. It should be easy to convince yourself that each of these are true using what we discovered about the group operation on three-element groups. Therefore $f$ is an isomorphism.
To test your understanding, try answering the following question: If we had defined $f$ to be $f(e_1)=e_2$, $f(x_1)=y_2$, and $f(y_1)=x_2$, would it still be an isomorphism?

As the comments hint, once you learn about cyclic groups and the order of groups, you can come up with a simpler (but more sophisticated) argument that proves a much more general version of this problem. 
