Groups of Order 8

Theorem
Let $G$ be a group of order 8.

Then $G$ is isomorphic to one of the following:


 * $\Z_8, \Z_4 \oplus \Z_2, \Z_2 \oplus \Z_2 \oplus \Z_2, D_4, Q_4$

where:


 * $\Z_n$ is the cyclic group of order $n$
 * $D_4$ is the dihedral group of order $8$
 * $Q_4$ is the dicyclic group of order $8$.

Proof
The abelian cases are handled by Abelian Group Factored by Prime/Corollary.

Suppose $G$ is non-abelian.

By Lagrange's theorem the order of non-identity elements in $G$ is either $2$, $4$ or $8$.

Aiming for Contradiction, suppose that there exists an order 8 element.

Then $G$ is generated by this element.

So $G$ is by definition cyclic.

But Cyclic Group is Abelian, contradicting the assumption that $G$ is non-abelian.

So there is no order 8 element.

By Boolean Group is Abelian, there exists at least one order 4 element in $G$.

Let it be denoted by $a$.

Let the subgroup generated by $a$ be denoted by $A$.

By Lagrange's theorem there are two cosets in $G$: $A$ and $G \setminus A$.

Pick $b \in G \setminus A$.

Then $\{a, b\}$ is a generating set of $G$.

Now we consider how $a$ and $b$ interact with each other.

Consider the element $x = b a b^{-1}$.

By Subgroup of Index 2 is Normal, $b A b^{-1} = A$.

So $x \in A$.

By Order of Conjugate Element, only possible choices are $x = a$ or $x = a^3$.

If $x = a$, then $a$, $b$ commutes.

Since $\{a, b\}$ generates $G$, this makes $G$ an abelian group, which is a contradiction.

So $b a b^{-1} = a^3$.

It suffices to consider the order of $b$.


 * If $\left\vert b \right\vert = 2$, then $G \cong D_4$.


 * If $\left\vert b \right\vert = 4$, then $G \cong Q_4$.

Also see

 * Classification of Groups of Order up to 15
 * Groups of Order 12