> [!NOTE] Theorem (Classification of Groups of Order 6) > Let $G$ be a [[Finite Group|finite group]] of order $6.$ Then $G$ is [[Homomorphisms of groups|isomorphic]] to either the [[Cyclic Group|cyclic group]] $C_{6}$ or the [[Dihedral Group|dihedral group]] $D_{6}.$ ###### Proof \[MA268\] Let $G$ be a group of order $6$. If $G$ has an element of order $6$ then $G \cong C_{6}$. Suppose otherwise that $G$ has no element of order $6$. Then by [[Lagrange's theorem (on Finite Groups)|Lagrange's theorem]], every element of $G$ has order $3$ or $2$. If $G$ no element of order $3$, then every element has order $2$ and so using [[Group of Exponent 2 is a Power of 2nd Cyclic Group|order of group of exponent 2]], $\#G = 2^n$ for some $n\in \mathbb{N}^+$ which is contradiction. Hence $G$ must have an element $b$ of order $3$. Applying Lagrange's theorem again yields $[G:\langle b \rangle]=6/3=2$. Since [[Subgroup of Index 2 is Normal|subgroups of index 2 are normal]], $\langle b \rangle$ is a normal subgroup of $G$. Let $a\in G-H$. Since $\#(G/H)=2$ we have $a^2H=(aH)^2=H$, so using [[Necessary Condition for Equality of Cosets|necessary condition for equality of cosets]] we have $a^2\in H$. If $a^2=b$ or $b^2$ then $a$ has order $6$ giving a contradiction. Hence $a^2 = \text{id}$. Consider $aba^{-1}$. As $\langle b \rangle$ is normal, this belongs to $H$. Thus $aba^{-1}=\text{id},b$ or $b^2$. If $aba^{-1}=\text{id}$ then $ab=a$ so $b=\text{id}$ giving a contradiction. If $aba^{-1}=b$ then $ab=ba$. We now check that $ab$ has order $6$ giving a contradiction. Thus $aba^{-1}=b_{2}=b^{-3}$. Note that the elements $a,b$ satisfy $b^3=a^2=\text{id}$ and $aba=b^{-1}$. By [[Universal Property of Group Presentations|fundamental theorem of group presentations]], there exists a homomorphism $\phi:D_{6}\to G$ such that $\phi(r)=b$ and $\phi(s)=a$. Recall that $D_{6}=\{ 1,r,r^2, s, sr, sr^{2} \}$and $G=\langle b \rangle \cup s \langle b \rangle = \{1, b, b^2, a, ab,ab^2 \}.$Clearly $\phi$ is surjective. Moreover, as $\# D_{6} = \#G= \#\text{Im}(\phi)$ the map $\phi$ is injective. Hence $\phi$ is an isomorphism. $\square$ Note that we can deduce the existence of an order $3$ element using Cauchy's theorem (see below). ###### Proof Let $G$ be a group of order $6$. If $G$ has an element of order $6$ then $G \cong C_{6}$. Suppose otherwise that $G$ has no element of order $6$. By [[Finite Group has Elements whose Orders are Each of The Prime Factors of Group Order (Cauchy's Theorem)|Cauchy's theorem]], we know that $G$ has an element of order $3$ (and also order $2$). Proceed as above.