Trying to show that any group of order four is either cyclic or isomorphic to V

Question

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?

Answer 1

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.

Answer 2

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 $c$.

Answer 3

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.