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. 
 A: Let's work on $G_1$ and assume $c=e_{G_1}$. So $$ac=a, bc=b,cc=c$$ But what about $ab$. It is either $a$, $b$ or $c$. If $ab=a=ac$, since $G_1$ is a group so we can cancel $a$ to have $b=c$ which is wrong because $b\neq c$. The same short story is when we assume $ab=b$, so we just have $ab=c$. This means that $$a=b^{-1}$$ and vise versa. Now what about $aa$? Is it equal to $a$, $b$ or $c$? We have $$aa=a\to a=c\\ aa=b\to aa=a^{-1}\to a^3=c\\ aa=c\to a=a^{-1}=b$$ The first one and the last one is clearly wrong so we just get $a^3=c$. Hence our group $G$ is change to the following form: $$G_1=\{c,a,a^{-1}\}$$ in which $a^3=c$. Now do the same way for $G_2$. It gives us: $$G_2=\{e,d,d^{-1}\}$$ What are the differences between these two groups? Just changing the alphabet $a\to d$ and vice versa? So, there are no differences between them and they are really the same things.
A: 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. 
A: HINT: Look at the Cayley table(s). Play some suduko.
(Doing it this way is no different from the standard proof that there is a single group of order two, up to isomorphism.)
A: $a$ can only have order $3$ and the same is true for $d$. Then $\left\{ a,b,c\right\} =\left\{ a,a^{2},a^{3}=c\right\} $ (so $a^{2}=b$)
and $\left\{ d,e,f\right\} =\left\{ d,d^{2},d^{3}=e\right\} $ (so
$d^{2}=f$) and the mapping $a\mapsto d$ , $(b=a^{2}) \mapsto (d^{2}=f)$, $c\mapsto e$
is an isomorphism.
A: Notice that any group with $3$ elements consists of just one generator, the inverse of this generator, and a neutral element (for otherwise it would not be a group). Therefore, there must exist a one-to-one correspondence between any two such groups.
A: Write down a map of the sets, then prove that it is an isomorphism.  Bijectivity should follow from your construction.  So prove that it is a homomorphism.  Hint:  Order of the elements should help.
