Any two groups of three elements are isomorphic - Fraleigh p. 47 4.25(b) The answer has no details. Hence maybe the answer is supposed to be quick. But I can't see it?
Hence I took two groups. Call them $G_1 = \{a, b, c\}, G_2 = \{d, e, f\}$.
Then because every group has an identity, I know $G_1, G_2$ has one each.
Hence WLOG pick $c$ as the identity in $G_1$. I want to match letters so pick $e$  as the identity in $G_2$.
Now we have $G_1 = \{a, b, \color{magenta}{c}\}, G_2 = \{d, \color{magenta}{e}, f\}$.
I know every group has an inverse. But how do I apply this to $G_1, G_2$ to simplify them?
And to prove $G_1, G_2$ are isomorphic, how do I envisage and envision what the isomorphism is?
Update Dec. 25, 2013 (1). Answer from B.S. Why does $ab = b$ fail? 
$\begin{align} ab & = b \\ & = bc \end{align}$. What now?
(2.) Do I have to do all the algebra work for $G_1$ for $G_2$? Is there some smart answer?
Update Jan. 8, 2014 (1.) I'm confounded by drhab's comment on Dec. 30 2013. Is drhab saying: Even if the domain's identity is the $\color{green}{third}$ letter $\color{magenta}{c}$  but the codomain's identity is the second letter $\color{magenta}{e}$, $d^{\huge{\color{green}{3}}} = e$ anyways? Hence I should've chosen the $\color{green}{third}$  letter in the codomain as the identity too?
What else is drhab saying about this?
(2.) I don't understand drhab's comment on Dec. 28 2013. Why refer to commutativity? It's not a group axiom? And what are the binary operations? 
Update: I didn't realize this before, but by dint of Martin Sleziak's comment, this question is just a sepcial case of Fraleigh p. 63 Theorem 6.10 = Pinter p. 109-111 Theorem 11.1. 

Let $p$ be a prime and $G$ a groups of order $p$. If $a\in G$ is not the neutral element show that the homomorphism $\mathbb Z\to G$ given by $1\mapsto a$ induces an isomorphims $\mathbb Z/p\mathbb Z\to G$. Conclude that any two groups of order $p$ are isomorphic. 
