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?
 A: By Lagrange's theorem, we must have that the order of $a$ is either $2$ or $4$. But if it were $4$ the group would be cyclic, so the order of $a$ is $2$, i.e. $a^2 = 1$. So if you assume $ab = 1$, you get $a = a1 = aab = a^2 b = b$, a contradiction (the starting assumption is that $G$ has four different elements, of course). So $ab$ cannot be $1$. It cannot be $a$ either, because $ab = a = a1 \implies b = 1$, and similarly it cannot be $b$. So it has to be $c$.
A: Here's a 'non-Lagrange' argument:
Pick an element $a \neq 1$ and consider the sequence $a$, $a^2$, $a^3$, $a^4$, $a^5$.  Since $G$ has only four distinct elements, these are not all distinct; there must be $i$, $j$ such that $a^i$=$a^j$.  Multiplying by the inverse of $a$ repeatedly if necessary, we may assume $i$=1.  Now then:
a) $a$=$a^2$ --> 1=$a$ --> CONTRADICTION.
b) $a$=$a^4$ --> $a^3$=1 --> 1, $a$, and $a^2$ are distinct elements of $G$.  There is one more element $b$ of $G$ different from all of these.  But then $b$, $ba$, and $ba^2$ are distinct elements of $G$.  Since 1, $a$, $a^2$, and $b$ exhausts $G$, $ba$ and $ba^2$ must each be one of these; but any such equality must imply another which contradicts a previous assumption.  Example: 1=$ba$ implies $a^2$=($ba$)$a^2$=$ba^3$=$b$ --> CONTRADICTION.
Therefore the only possibilities not yet excluded are $a$=$a^3$ and $a$=$a^5$.  A little work will show that each of these yields a group (try to compute the product tables).  The first yields the non-cyclic group, the second yields the cyclic group.
A: 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.
