Groups of Order 8

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

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


 * $\Z_8$
 * $\Z_4 \oplus \Z_2$
 * $\Z_2 \oplus \Z_2 \oplus \Z_2$
 * $D_4$
 * $\Dic 2$

where:


 * $\Z_n$ is the cyclic group of order $n$
 * $D_4$ is the dihedral group of order $8$
 * $\Dic 2$ is the dicyclic group of order $8$.

Proof
The abelian cases are handled by the corollary to Abelian Group Factored by Prime.

Let $G$ be non-abelian.

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

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 Non-Abelian Order 8 Group has Order 4 Element, there exists at least one order $4$ element in $G$.

Let it be denoted by $a$.

Let $A$ denote the subgroup generated by $a$.

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

Let $b \in G \setminus A$.

Then $\set {a, b}$ is a generator 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, the only possible choices are $x = a$ or $x = a^3$.

If $x = a$, then $a$ and $b$ commute.

Since $\set {a, b}$ 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 $\order b = 2$, then $G \cong D_4$.


 * If $\order b = 4$, then $G \cong \Dic 2$.

Also see

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