Showing there is a unique group table for $\{1, a,b,c\}$ such that there is no element of order $4$.

Question

Assume $G = \{1, a,b,c\}$ is a group of order $4$ with identity $1.$ Assume also that $G$ has no elements of order $4$. Show that there is a unique group table for $G$. Also show that $G$ is abelian.

If $G$ is abelian, then the group table matrix must be symmetric. How can I introduce a binary function and show it? I am new in this field, so I am not so familiar. I have proved many other exercises, but it is a little tough (for me).

Can you please help?

Edit: I know every element has order $\leq 3$ , but I do not understand how I will proceed.

Answer 1

Just use juxtaposition for the binary operation.

Further, here is a . . .

Hint: If $a^2=b$, then $ab=a^3\neq 1$ by Lagrange's Theorem.

The group is (isomorphic to) $\Bbb Z_2\times \Bbb Z_2$.

Answer 2

The order of the elements must divide the order of the group. Since $|G|=4$ and there's no elements of order $4$ all elements must have order $2$ or order $1$ (clearly the latter must be the identity).

The element $ab$ must be in the group. $ab\not = e$ because $a^2=e$ and this would imply $a=b$. Also $ab\not = a$ because then $b=e$ and $ab\not = b$ because then $a=e$. In other words $ab=c$.

Symmetric argument will also give that $ba=c$. Similarly $ac=ca=b$ and $bc=cb=a$.

We completely calculated all the products in this group.

$$\left[\begin{array}{c|cccc}* & \textbf{1} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{1} & 1 &a &b&c \\ \textbf{a} &a&1&c&b \\ \textbf{b} &b&c&1&a\\ \textbf{c}&c&b&a&1\end{array}\right]$$