Group Theory

We classify all groups with at most eight elements. Recall groups of prime order are cyclic, so we need only focus on the cases $\mid G\mid =\mathrm{4,6,8}$. 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 $\mid G\mid$ is a power of 2.

Proof: Clearly true for $\mid G\mid =2$. 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 $ab\ne 1$ (otherwise $a=b^{-1}=b$) and $1=(ab{)}^2=a(ba)b$ which implies $ba=a^{-1}{b}^{-1}=ab$. 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 $e_i\in \{\mathrm{0,1}\}$. Then $G=⟨a_1⟩\times ...\times ⟨a_n⟩$ and $\mid G\mid =2\times ...\times 2=2^n$

Now we can classify the groups up to order eight:

• $\mid G\mid =4$: 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").

• $\mid G\mid =6$: 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 $1,a,a^2,b,ab,a^{2}b$ 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 $1=ab$ or $1=a^{2}b$, both of which are contradictions. Hence $b^2=1$. Next we determine which element is equal to $ba$. The only possible choices are $ab$ or $a^{2}b$. If $ba=ab$, then $G$ is abelian, but then $(ab{)}^2=a^2$ and $(ab{)}^3=b$ implying that $ab$ has order 6, a contradiction. Thus $ba=a^{2}b$, implying $(ab{)}^2=1$. We have defining relations $a^3=b^2=(ab{)}^2=1$. 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).

• $\mid G\mid =8$: 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 $1,a,a^2,a^3,b,ab,a^{2}b,a^{3}b$. 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 $ba$ must be equal to one of the last three elements. If $ba=ab$ then the group is abelian and we end up with the aforementioned ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$. If $ba=a^{2}b$, then we have $b^{-1}{a}^{2}b=a$. Upon squaring, we derive the contradictory $a^2=1$. So we must have $ba=a^{3}b$, that is, $(ab{)}^2=1$. The defining relations are $a^4=b^2=(ab{)}^2=1$, 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 $ba=ab$ then the group is abelian and again we wind up with the group ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$. If $ba=a^{2}b$ we have $ba=b^3$, which is a contradiction because it implies $a=b^2=a^2$. Thus we must have $ba=a^{3}b$. Then we get a group with the defining relations $a^4=1,a^2=b^2,\mathrm{ba}=a^{3}b$, which is known as the quaternion group. To verify associativity, one can show it is isomorphic to the group generated by the matrices

  $$\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right),\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$$

  or

  $$\left(\begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ -1& 0& 0& 0\\ 0& -1& 0& 0\end{array}\right)$$

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