> [!NOTE] Theorem (Classification of Groups of Order 6) > Let $G$ be a [[Finite Group|finite group]] of order $8.$ Then $G$ is [[Homomorphisms of groups|isomorphic]] to one of $C_2 \times C_{2} \times C_{2}, \quad C_{4} \times C_{2}, \quad C_{8}, \quad D_{8}, \quad Q_{8}$ where $C_{n}$ is the [[Cyclic Group|cyclic group]] of order $8$, $\times$ represents [[Cartesian Product|cartesian product]] and $D_{8}$ is the [[Dihedral Group|dihedral group]] of order $8$ and $Q_{8}$ is the [[Quaternion Group|quaternion group]] . ###### Proof \[MA268\] Let $G$ be a group of order $8$. By [[Lagrange's theorem (on Finite Groups)|Lagrange]], every element has order $1,2,4$ or $8$. If $G$ has an element of order $8$ then $G \cong C_{8}$. If $G$ has exponent $2$, then applying the [[Group of Exponent 2 is a Power of 2nd Cyclic Group|fundamental theorem of finite Abelian groups]] yields $G\cong C_{2} \times C_{2} \times C_{2}$. Now suppose $G$ has an element $u$ of order $4$ and no element of order $8$. Let $H=\langle u \rangle$. Since $[G:H]=2$ and [[Subgroup of Index 2 is Normal|index 2 subgroups are normal]] $H$ is normal in $G$. Choose $w\not \in H$. Since the quotient group $G/H$ has order $2$, $(wH)^2=H$ and so $w^2\in H$. Thus $w^2 =\text{id},u,u^2$ or $u^3$. However, if $w^2 = u$ then $w$ has order $8$ giving a contradiction. Thus $w^2=\mathrm{id}\quad\mathrm{or}\quad w^2=u^2.$ Also as $u\in H$ and $H$ is normal, we have $wuw^{-1}\in H$, and so $wuw^{-1}=\text{id},u,u^2,u^3$. But if $wuw^{-1}=\text{id}$ then $u=\text{id}$ giving a contradiction. Moreover, if $wuw^{-1}=u^2$, then $\text{id}=u^4=(wuw^{-1})^2=wuw^{-1}\cdot wuw^{-1}=wu^2w^{-1}$ so $u^2=\text{id}$ giving a contradiction. Hence we have two possibilities $wuw^{-1}=u\quad\mathrm{or}\quad wuw^{-1}=u^3.$ We deal with possibilities separately. **Case 1**: $w^2= id$ and $wuw^{-1}=u.$ The latter identity tells us that $w$ and $u$ commute: $wu=uw.$ Moreover, $\langle w\rangle\cap\langle u\rangle=\{$id$\}.$ Then $G=H\cup wH=\langle u,w\rangle\cong\langle u\rangle\times\langle w\rangle\cong C_4\times C_2.$ **Case 2**: $w^2=u^2$ and $wuw^{-1}$. Again the latter identity tells us that $w$ and $u$ commute: $wu=uw$. Let $v=wu$. Then $v \neq \text{id}$ since $w\not \in H=\langle u \rangle$. But $v^2=w^2u^2=u^4=\text{id}$. Moreover $vu=uv$. We're now reduced to Case 1 with $v$ in place of $w.$ **Case 3**: $w^2=$id and $wuw^-1=u^3=u^{-1}.$ Define $\phi:D_8\to G,\quad\phi(r)=u,\quad\phi(s)=w.$ Using the [[Universal Property of Group Presentations|fundamental theorem of group presentations]] this is a homomorphism. It is surjective as $G=$ $\langle u,w\rangle.$ Moreover, $\#D_8=\#G=8$ so the map must be an injection. Hence $\phi$ is an isomorphism, so $G\cong D_8.$ **Case 4**: $w^2= u^2$ and $wuw^-1=u^3=u^{-1}.$ We define a map $\phi:Q_8\to G,\quad a\mapsto u,\quad b\mapsto v,$ and similarly argue that it is an isomorphism. Hence $G\cong Q_8.$ $\square$