# Groups Up To Order Eight

We classify all groups with at most eight elements. Recall groups of prime order are cyclic, so we need only focus on the cases $$|G| = 4,6,8$$. We make use of the following:

Lemma: If each element $$1 \ne g\in G$$ is of order 2, then $$G$$ is abelian and isomorphic to $$\mathbb{Z}_2 \times ... \times \mathbb{Z}_2$$ and $$|G|$$ is a power of 2.

Proof: Clearly true for $$|G| = 2$$. Otherwise, let $$1 \ne a \ne b \in G$$. We have $$a^2 = b^2 = 1$$, that is $$a = a^{-1}, b=b^{-1}$$. Then $$a b \ne 1$$ (otherwise $$a = b^{-1} = b$$) and $$1 = (a b)^2 = a (b a)b$$ which implies $$b a = a^{-1} b^{-1} = a b$$. Thus $$G$$ is abelian.

Since $$G$$ is finite, it has a finite set of independent generators $$a_1,...,a_n$$. As $$G$$ abelian, we may write an element $$g \in G$$ in the form

$g = a_1^{e_1} ... a_n^{e_n}$

where each $$e_i \in\{0,1\}$$. Then $$G = \langle a_1 \rangle \times ... \times \langle a_n \rangle$$ and $$|G| = 2 \times ... \times 2 = 2^n$$

Now we can classify the groups up to order eight:

• $$|G| = 4$$: Each element (besides the identity) must have order 2 or 4. If $$a \in G$$ has order 4 it generates $$G$$ and we have $$G = \mathbb{Z}_4$$. Otherwise every element has order 2 and by the lemma we have $$G = \mathbb{Z}_2 \times \mathbb{Z}_2$$ (the four-group or quadratic group, sometimes denoted by $$V$$ after F. Klein’s "Vierergruppe").

• $$|G| = 6$$: If $$a \in G$$ has order 6 we have $$G = \mathbb{Z}_6$$. Otherwise all elements (besides the identity) have order 2 or 3. By the lemma, not all elements can have order 2 because 6 is not a power of 2. So let $$a$$ be an element of order 3, that is $$1,a,a^2$$ are distinct. Let $$b$$ be some other element in $$G$$. It can be verified that $$1,a,a^2,b,a b,a^2 b$$ must be distinct. In order to satisfy closure, $$b^2$$ must be one of these elements. The only possibilities are $$b^2 = 1,a$$ or $$a^2$$.

If $$b^2=a,a^2$$ we find that $$b$$ cannot have order 2, so it has order 3. Then $$1 = a b$$ or $$1 =a^2 b$$, both of which are contradictions. Hence $$b^2 = 1$$. Next we determine which element is equal to $$b a$$. The only possible choices are $$a b$$ or $$a^2 b$$. If $$b a = a b$$, then $$G$$ is abelian, but then $$(a b)^2 = a^2$$ and $$(a b)^3 = b$$ implying that $$a b$$ has order 6, a contradiction. Thus $$b a = a^2 b$$, implying $$(a b)^2 = 1$$. We have defining relations $$a^3 = b^2 = (a b)^2 = 1$$. We shall see later that this is indeed a group (associativity turns out to hold) because it is the symmetric group of degree 3 (which is isomorphic to the dihedral group of order 6).

• $$|G|=8$$: It turns out there are 3 abelian groups and 2 nonabelian groups. The three abelian groups are easy to classify: $$\mathbb{Z}_8, \mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$$.

The other groups must have the maximum order of any element greater than 2 but less than 8. Hence there exists an element of order 4, which we denote by $$a$$. All the others (besides the identity) have order 2 or 4. Let $$b$$ be an element not generated by $$a$$. Then we have the distinct elements $$1,a,a^2,a^3,b,a b,a^2 b,a^3 b$$. Now $$b^2$$ can only be one of the first four. But $$b^2 = a, a^3$$ imply $$b$$ is not of order 2 or 4, so we must have $$b^2 =1$$ or $$b^2= a^2$$.

Suppose $$b^2 = 1$$. Now $$b a$$ must be equal to one of the last three elements. If $$b a = a b$$ then the group is abelian and we end up with the aforementioned $$\mathbb{Z}_4 \times \mathbb{Z}_2$$. If $$b a = a^2 b$$, then we have $$b^{-1}a^2 b = a$$. Upon squaring, we derive the contradictory $$a^2 = 1$$. So we must have $$b a = a^3 b$$, that is, $$(a b)^2 = 1$$. The defining relations are $$a^4 = b^2 = (a b)^2 = 1$$, and this turns out to be the dihedral group of order 8, also known as the octic group.

The other possibility is $$b^2 = a^2$$. In this case, $$b$$ also has order 4. If $$b a = a b$$ then the group is abelian and again we wind up with the group $$\mathbb{Z}_4 \times \mathbb{Z}_2$$. If $$b a = a^2 b$$ we have $$b a = b^3$$, which is a contradiction because it implies $$a = b^2 = a^2$$. Thus we must have $$b a = a^3 b$$. Then we get a group with the defining relations $$a^4 = 1, a^2 = b^2, ba = a^3 b$$, which is known as the quaternion group. To verify associativity, one can show it is isomorphic to the group generated by the matrices

$\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

+ or

+

$\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}$

+ The quaternion group is a special case of a dicyclic group, groups of order $$4 m$$ given by $$a^{2m} = 1, a^m = (a b)^2 = b^2$$, and whose elements can be written $$1,a,...,a^{2m-1},b,a b,...,a^{2m -1}b$$. The square of elements not generated by $$a$$ is $$b^2$$.

Ben Lynn blynn@cs.stanford.edu 💡