----- > [!proposition] Proposition. ([[classification of groups of order 12]]) > The [[group]]s $G$ of order $12$ are, up to [[group isomorphism|isomorphism]], > - $C_{12}$ >- $C_{6} \times C_{2}$; >- $D_{6}$; >- $A_{4}$; >- $C_{3} \rtimes C_{4} =\langle x, y : x^{4}=e, y^{3}=e, xyx ^{-1} = y ^{-1} \rangle$ > [!proof]- Proof. ([[classification of groups of order 12]]) > Observe that $\begin{align} > 12 \\ > = & 2^{2} \cdot 3 \\ > = & 3^{1} \cdot 4, > \end{align}$and moreover $1 \text{ mod }2 = 1$, $3 \text{ mod } 2=1$, $1 \text{ mod } 3 = 1$, $4 \text{ mod }3=1$, so by [[the Sylow theorems|the first and third Sylow theorems]], $G$ may contain $\begin{align} > n_{2} \in & \{ 1,3 \} \ \ \text{2-Sylow subgroups of order 4} \\ > n_{3} \in & \{ 1,4 \} \text{ 3-Sylow subgroups of order 3}, > \end{align}$ > and all these intersect trivially (in general, [[subgroups of coprime order intersect trivially]]). > > Notice that it is impossible to have $n_{2}=3$ while $n_{3}=4$ because $1+3(3)+2(4)>12$. Thus $n_{2}=1$ or $n_{3}=1$, guaranteeing by the [[the Sylow theorems|second Sylow theorem]] that $G$ must have a [[normal subgroup|normal]] $2$-Sylow [[subgroup]] or a [[normal subgroup|normal]] $3$-Sylow [[subgroup]]; call it $N$. Because $NJ=JN$ for all [[subgroup]]s $J$ of $G$, $NJ$ is a [[subgroup]] for all [[subgroup]]s $J$ of $G$ by [[Frobenius product of subgroups|this property]]. By [[cardinality of frobenius product of subgroups]], we can say that if $J$ is a $p$-sylow subgroup of different order than $N$, $|NJ|=|N||J|=12=|G|$, and thus $NJ=G$. From this we conclude that $G$ is a [[external semi-direct product|semi-direct product]] $H \rtimes K$ where either $|H|=3$ or $|H|=4$. > > # Case 1: $|H|=3$ > Set $H:=\{ h^{0}, h^{1},h^{2} \}$ and $K=\{ k^{0}, k^{1}, k^{2}, k^{3} \}$. > > ## Case 1.1: $K \cong \mathbb{Z}_{4}$ > First, we must find all [[group homomorphism|homomorphisms]] from $\mathbb{Z}_{4} \to \text{Aut}(\mathbb{Z}_{3})\cong \mathbb{Z}_{2}$. This amounts to finding all [[group homomorphism|homomorphisms]] from $\{ 0,1,2,3 \} \to \{ 0, 1 \}$. Any map which sends $1 \mapsto 0$ is the trivial [[group homomorphism|trivial homomorphism]]: $f_{\text{trivial}}$ sends $1 \mapsto 0$, and hence $f_\text{trivial}(2)=f_\text{trivial}(1+1)=f_{\text{trivial}}(1)f_{\text{trivial}}(1)=0$; likewise $f_\text{trivial}(3)=0$. The only other option for a [[group homomorphism|homomorphism]] is some map $f$ which sends $1 \mapsto 1$; in this case $f(2)=f(1+1)=f(1)+f(1)=0$ and $f(3)=(f(2) + 1)=0+1=1$ so the map is completely determined by this property. Thus there are *$2$ homomorphisms total*, $f_\text{trivial}$ and $f$, to consider. > > ### Case 1.1.1: $\phi=f_\text{trivial}: \mathbb{Z}_{4} \to \text{Aut}(\mathbb{Z}_{3})$ > In this case the multiplication operation just becomes that of the [[direct product of groups|direct product]], and hence $G \cong C_{4} \times C_{3}$. $4$ and $3$ are [[relatively prime integers|coprime]], so $G \cong C_{12}$. > > ### Case 1.1.2: $\phi=f_{1 \mapsto 1}: \mathbb{Z}_{4} \to \text{Aut}(\mathbb{Z}_{3})$ > > This is a prospective group distinct from $C_{12}$. Thus, **case 1.1. yielded up to two groups of order 12.** > > ## Case 1.2: $K \cong C_{2} \times C_{2}$ > > Write $\text{Aut}(H)=\{ \id, \text{inv} \}$ where $\id$ is the [[identity map]] and $\text{inv}$ is the map which swaps the nontrivial elements in $H$. > > We must first find all [[group homomorphism|homomorphisms]] $f$ from $\{ e,a,b,ab \} \to \{ \id, \text{inv}\}$. > Clearly the [[group homomorphism|trivial homomorphism]] is one such, so suppose $f \neq f_{\text{trivial}}$. If $f(a)$ and $f(b)$ are $\id$, then $f(ab)$ is $\id$ too and $f=f_{\text{trivial}}$; hence we need to have $f(a)$ or $f(b)$ equal to $\text{inv}$. One option is $f(a) =\text{inv}$ and $f(b)=\id$ with $f(ab)=\text{inv}$ to maintain additivity. A similar option is setting $f(a)=\id$ and $f(b)=\text{inv}$. A third option is setting $f(a)=\text{inv}=f(b)$ and $f(ab)=\id$. So there are *$4$ homomorphisms total.*, the trivial one and then three which are characterized by which nontrivial element they send to $\text{id}$. Write them as $f_{a},f_{b},f_{ab}$. > > ### Case 1.2.1: $\phi=f_{\text{trivial}}: C_{2}\times C_{2} \to \text{Aut }H$ > In this case the multiplication operation just becomes that of the [[direct product of groups|direct product]], and hence $G \cong C_{3} \times C_{2} \times C_{2} \cong C_{6} \times C_{2}$. > > ### Case 1.2.2: $\phi_{}=f_{a}:C_{2} \times C_{2} \to \text{Aut }H$ > > We have $f_{a}=f_{b} \circ \mu$, where $\mu$ the [[permutation]] given by $\mu(a)=b$, $\mu(b)=a$, and $\mu(ab)=ab$. ($\mu \in \text{Aut}(C_{2} \times C_{2})$ by [[automorphism|this result]]). Now we can apply 3c with $\psi=\psi'=\mu'=\id$ to ascertain that $f_{a}$ and $f_{b}$ yield isomorphic groups. > > ### Case 1.2.3: $\phi_{}=f_{b}: C_{2} \times C_{2} \to \text{Aut }H$ > > We have $\phi_{b}=\phi_{ab}\circ \mu$, where $\mu$ is the [[permutation]] given by $\mu(ab)=b$, $\mu(b)=ab$, $\mu(a)=a$. Now we can apply 3c with $\psi=\psi'=\mu'=\id$ to ascertain that $f_{b}$ and $f_{ab}$ yield isomorphic groups. > > ### Case 1.2.4: $\phi_{}=f_{ab}: C_{2} \times C_{2} \to \text{Aut }H$ > Putting together the two cases above tells us that $f_{a},f_{b},$ and $f_{ab}$ all yield [[group isomorphism|isomorphic groups]]. > > Thus, **case 1.2. yielded up to two additional groups of order 12.** > > # Case 2: $|H|=4$ > ## Case 2.1: $H \cong \mathbb{Z}_{4}$ > We need to find all [[group homomorphism|homomorphisms]] from $\{ 0,1,2 \} \to \text{Aut}(C_{4}) \cong \mathbb{Z}_{2}$ (that latter isomorphism holds because the [[generating set of a group|generator]] $1$ needs to go to an element of order $4$ — options are $a$ and $a^{3}$). This amounts to finding all [[group homomorphism|homomorphisms]] from $\{ 0,1,2 \} \to \{ 0,1 \}$. Any such [[group homomorphism|homomorphism]] is characterized by where it takes the generator $1$; but if $1 \mapsto 1$ then the order of its image doesn't [[divides|divide]] $3$ (see [[group homomorphisms preserve structure]]). So there is *one homomorphism in total* of these: the trivial [[group homomorphism|homomorphism]] that sends every source element to the target's identity. > > ### Case 2.1.1: $\phi=f_\text{trivial}:\mathbb{Z}_{3} \to \text{Aut}(\mathbb{Z}_4)$ > In this case the multiplication operation just becomes that of the [[direct product of groups|direct product]], and hence $G \cong C_{3} \times C_{4}\cong C_{12}$. > > Thus, **case 2.1. yielded no additional groups of order 12.** > > > ## Case 2.2: $H \cong C_{2} \times C_{2}$ > We need to find all [[group homomorphism|homomorphisms]] $f$ from $\{ 0,1,2 \} \to \text{Aut}(C_{2} \times C_{2}) \cong S_{3}$ (see [[automorphism|here]] for a proof of the latter isomorphism). > > The only [[subgroup]] of $S_{3}$ of [[order of an element in a group|order]] $3$ is $\{ e,\tau,\tau^{2} \}$. $1$ has order $3$ in $\{ 0,1,2 \}$, therefore [[group homomorphisms preserve structure|we need]] $|f(1)|\ \b | \ 3$, i.e., $|f(1)|=1$ or $|f(1)|=3$. > > If $|f(1)|=1$ then $f$ is the [[group homomorphism|trivial homomorphism]][[group homomorphism|]], and we just get that $G$ is the [[direct product of groups|direct product]] $H \times K$. > > If $|f(1)|=3$ then we get that $f(1)$ generates $\{ e,\tau,\tau^{2} \}$ since [[subgroup#^d58546|this is the only order-3 subgroup of]] $S_{3}$. Nothing new happens if we trivially send $1 \mapsto e$. If we send $1 \mapsto \tau$, then if we also send $2 \mapsto \tau^{2}$ we have a new homomorphism. If we send $1 \mapsto \tau^{2}$, then if we also send $2 \mapsto (\tau^{2})^{2}=\tau$ we have yet another homomorphism. So, there are *three homomorphisms total*. > > ### Case 2.2.1: $\phi=f_{\text{trivial}}:\mathbb{Z}_{3} \to \text{Aut}(C_{2} \times C_{2})$ > In this case the multiplication operation just becomes that of the [[direct product of groups|direct product]], and hence $G \cong C_{3} \times C_{2} \times C_{2}$. > > ### Case 2.2.2: $\phi{^{(\tau)}}=f_{1 \mapsto \tau}: \mathbb{Z}_{3} \to \text{Aut}(C_{2} \times C_{2})$ > > We need to find $\mu \in \text{Aut}(\mathbb{Z}_3)$ such that $\phi^{(\tau)}=\phi^{(\tau^{2})}\circ \mu$. The only option to pick is $\mu$ that swaps nontrivial elements, and indeed we see this map satisfies the equality. Thus the group from $\phi^{(\tau)}$ is isomorphic to that from $\phi^{(\tau^{2})}$ (when we apply 3c, taking $\psi=\psi'=\mu'=\id$). > > ### Case 2.2.3: $\phi^{(\tau^{2})}=f_{1 \mapsto \tau^{2}}: \mathbb{Z}_{3} \to \text{Aut}(C_{2} \times C_{2})$ > See case 2.2.2. above. > > Thus, **case 2.2. yielded up to one additional group of order 12.** > > In summary, these cases show that, up to [[group isomorphism|isomorphism]], there are at most $5$ [[group]]s of order $12$. So, we will hence provide $5$ groups of order 12 which are not isomorphic and that will complete the classification: > - $C_{12}$ > - $C_{6} \times C_{2}$ > - $D_{6}$ > - $A_{4}$ > - $C_{3} \rtimes C_{4} \cong$ $\langle x, y : x^{4}=e, y^{3}=e, xyx ^{-1} = y ^{-1} \rangle$ > > > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```