What Are the Group Tables for a Group of Four Elements?

A group with four elements has one identity and three nonidentity elements. Up to isomorphism, there are exactly two possibilities: the cyclic group $C_4$ and the Klein four-group $V_4$. Their multiplication tables are different, even though both groups are abelian.

In the tables below, the entry in a row and column is the product of the row element followed by the column element. For example, the entry in row $a$ and column $b$ is $ab$.

The cyclic group $C_4$

The cyclic group has an element $a$ of order four, meaning $a^4=e$ and no smaller positive power of $a$ is the identity. Its elements are $e,a,a^2,a^3$, and every product can be reduced using $a^4=e$.

$\cdot$	$e$	$a$	$a^2$	$a^3$
$e$	$e$	$a$	$a^2$	$a^3$
$a$	$a$	$a^2$	$a^3$	$e$
$a^2$	$a^2$	$a^3$	$e$	$a$
$a^3$	$a^3$	$e$	$a$	$a^2$

For instance, the entry in row $a^2$ and column $a^3$ is $a^5=a$, since $a^4=e$. The inverse pairs are $e^{-1}=e$, $a^{-1}=a^3$, and $(a^2)^{-1}=a^2$.

The Klein four-group $V_4$

In the Klein four-group, every nonidentity element squares to the identity. Write its elements as $e,a,b,c$, with $c=ab$. Then $ab=c$, and the multiplication rules imply $ac=b$ and $bc=a$. The table is

$\cdot$	$e$	$a$	$b$	$c$
$e$	$e$	$a$	$b$	$c$
$a$	$a$	$e$	$c$	$b$
$b$	$b$	$c$	$e$	$a$
$c$	$c$	$b$	$a$	$e$

For example, $a\cdot c=b$. Since $c=ab$ and $a^2=e$, this follows from $a(ab)=(a^2)b=b$. Every element is its own inverse, and the table is symmetric because the group is abelian.

A common model for this group is $\mathbb{Z}_2\times\mathbb{Z}_2$, with componentwise addition modulo 2. Its elements are $(0,0),(1,0),(0,1),(1,1)$; each nonzero element added to itself gives $(0,0)$.

Why these are the only two tables

Let $G$ be any group with four elements. The order of an element divides the order of the group, so each nonidentity element has order two or four.

If some element $a$ has order four, then $e,a,a^2,a^3$ are all the elements of $G$. The group is cyclic, and its table must be the $C_4$ table above.

Otherwise, every nonidentity element has order two. Choose two distinct nonidentity elements $a$ and $b$. Their product cannot be $e$, $a$, or $b$: those possibilities would force one of $a,b$ to be the identity or force $a=b$. Thus $ab$ is the third nonidentity element; call it $c$. Also, $(ab)^{-1}=b^{-1}a^{-1}=ba$, while $ab=c$ is its own inverse. Hence $ba=ab$, and the products are exactly those in the $V_4$ table. This proves there are no other groups of order four up to isomorphism.

Common mistakes

• Confusing order four with four generators. The order of a group is its number of elements; a group of order four need not be cyclic.
• Assuming every group of order four is $C_4$. The Klein four-group is not cyclic: none of its elements generates all four elements.
• Treating the tables as different because of labels. Renaming elements does not change a group’s structure. The key distinction is element order: $C_4$ has an element of order four, while $V_4$ does not.
• Forgetting the product order convention. In a nonabelian group, row-times-column and column-times-row can differ. They happen to agree in both groups shown here.