<s>[INST] Trying to show that any group of order four is either cyclic or isomorphic to V I know the question has already been asked. But I have trouble with the answer. 
Having a non-cyclic group $\,G=\{1,a,b,c\}\,$, how can I show that $ab=c$? 
In my attempt, I assume that $ab=1$, and then $c^2=1$. And I see no problem with that. What am I missing?
UPD: is it possible to do without Lagrange's theorem?
 [/INST] We surely have $a^2\ne a$. So $a^2=1$, $a^2=b$ or $a^2=c$. If $a^2\ne1$, it's not restrictive to assume $a^2=b$. So we can start building the Cayley table:
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b &   &   \\
b & b &   &   &   \\
c & c &   &   & 
\end{array}
$$
We can see that $ab=c$ (otherwise $ab=1$ and we'd remain with $ac=c$, a contradiction); thus $ac=1$. Similarly, $ba=c$ and $ca=1$.
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & c & 1 \\
b & b & c &   &   \\
c & c & 1 &   & 
\end{array}
$$
Now we are forced to $b^2=1$ and $bc=a$; similarly, $cb=a$; finally, $c^2=b$:
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & c & 1 \\
b & b & c & 1 & a \\
c & c & 1 & a & b
\end{array}
$$
Thus we see that $a^0=1$, $a^2=b$, $a^3=c$; hence the group is cyclic.
In order not to have a cyclic group we therefore need $a^2=1$:
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 &   &   \\
b & b &   &   &   \\
c & c &   &   & 
\end{array}
$$
We then must have $ab=c$ and $ac=b$; similarly, $ba=c$ and $ca=b$:
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c &   &   \\
c & c & b &   & 
\end{array}
$$
If $b^2=a$, we'd be in the same situation as before, because then $b^3=c$ and the group would be cyclic. So, for a non cyclic group we must have $b^2=1$ and $bc=a$. Now we can complete the Cayley table:
$$
\begin{array}{c|cccc}
  & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & 1 & a \\
c & c & b & a & 1
\end{array}
$$
which shows the group is the Klein group.
</s>