Groups of Order 4 – Nilradical

In the typical study of finite groups students begin by learning about cyclic groups. For simplicity, we usually write the cyclic groups in terms of the integers mod $n$ .

$\mathbb{Z}_2=\{0,1\} $$

$\mathbb{Z}_3=\{0,1,2\}$

When we come to groups of order $4$ we encounter a new situation. We get the usual cyclic group of order $4$

$\mathbb{Z}_4=\{0,1,2,3\}$

However, the new situation we encounter is that $4$ is not prime. Thus, we can ask the question; what happens if we take the direct product of cyclic groups of order $2$ ?

$\mathbb{Z}_2\,\times\,\mathbb{Z}_2=\{(0,0),(0,1),(1,0),(1,1)\}$

This is indeed a group of order $4$ , but what is the operation and is it isomorphic to $\mathbb{Z}_4$ ? We could try to define what seems like the natural operation where $a*b=(a_1+b_1,a_2+b_2)$ . Obviously, this definition leads to $(0,0)$ being the identity element. Something curious happens when we square any element.

$(0,1)*(0,1)=(0+0,1+1)=(0,0)$

$(1,0)*(1,0)=(1+1,0+0)=(0,0)$

$(1,1)*(1,1)=(1+1,1+1)=(0,0)$

We immediately see that this group is not isomorphic to the cyclic group because no single element can generate the entire group. It’s easy to see that the Cayley table for this group looks like this

*	$(0,0)$	$(0,1)$	$(1,0)$	$(1,1)$
$(0,0)$	$(0,0)$	$(0,1)$	$(1,0)$	$(1,1)$
$(0,1)$	$(0,1)$	$(0,0)$	$(1,1)$	$(1,0)$
$(1,0)$	$(1,0)$	$(1,1)$	$(0,0)$	$(0,1)$
$(1,1)$	$(1,1)$	$(1,0)$	$(0,1)$	$(0,0)$

We call this group the Klein 4 Group and denote it as $V_4=\mathbb{Z}_2\,\times\,\mathbb{Z}_2$ .

If we rename the elements like this

$(0,0)=e$

$(0,1)=a$

$(1,0)=b$

$(1,1)=c$

we can easily see that $\langle a\rangle\simeq\mathbb{Z}_2, \langle b\rangle\simeq\mathbb{Z}_2,$ and $\langle c\rangle\simeq\mathbb{Z}_2$ .

We leave it as an exercise to the reader to verify that there are only two groups of order four, up to isomorphism. That is $\mathbb{Z}_4$ and $V_4$ .

Leave a Reply Cancel reply