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
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 fourgroup 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
+
+ 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 \(1,a,...,a^{2m1},b,a b,...,a^{2m 1}b\). The square of elements not generated by \(a\) is \(b^2\).