Groups of Order 8

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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.

$\Box$


Let $G$ be non-abelian.

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

Aiming for a 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 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 equals Order of 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 $G \cong \Dic 2$.

$\blacksquare$


Also see


Sources

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