Exercise 1.1.36 (Table of the non cyclic group of order $4$)

Proof. Let $a\in G$. Consider the map

Then ${\phi }_a$ is injective: By the cancellation laws, for all $x,y\in G$,

Moreover ${\phi }_a$ is surjective: If $y$ is any element in $G$, put $x=a^{-1}y$. Then ${\phi }_a(x)={\phi }_a(a^{-1}y)=y$.

Therefore ${\phi }_a$ is bijective.

If $G$ is a finite group, since ${\phi }_a(x)$ take the values of the line relative to $a$ in the table of $G$, this row of the table contains each element of $G$ once and only once.

Replacing ${\phi }_a$ by ${\psi }_b:x\mapsto \mathit{xb}$, we show similarly that every column of the table contains each element of $G$ once and only once.

(A table with these properties is called a latin square. Since the order of every element divides $|G|=4$ by the Lagrange’s Theorem (see Ex.1.7.19), and since by hypothesis $G$ has no element of order $4$, we conclude that every element $x\ne 1$ is of order $2$. Therefore

If we use $1\cdot x=x\cdot 1=x$ for every $x\in G$, and complete the square as a Sudoku, using the fact that this is a latin square, we obtain the table

so there is a unique table for $G$.

We observe on this table that $\mathit{xy}=\mathit{yx}$ for all $x,y\in G$, so $G$ is abelian. □