# Classifying groups of order $12$.

I want to classify all groups of order $$12$$.

Let $$G$$ be a group with $$|G|=12$$. Then $$n_3=1$$ or $$4$$.

1. If $$n_3=4$$ then we have $$|G:\langle x \rangle |=4$$ where $$\langle x\rangle$$ is a Sylow $$3-$$ subgroup (not normal in $$G$$) so we have a homomorphism $$r:G\to S_4$$ with $$ker(r)\subseteq \langle x\rangle$$ and $$ker(r)\lhd G\Rightarrow ker(r)=\{1\}$$ so $$G$$ is embedded in $$S_4$$ and has oredr $$12$$ hence $$G\cong A_4$$.
2. If $$n_3=1$$ then we have a unique Sylow $$3-$$ subgroup $$P=\langle x\rangle$$ and let $$H$$ a Sylow $$2-$$subgroup of $$G$$. Then $$G= P\rtimes_u H$$ where $$u:H\to Aut(P)$$ and $$Aut(P)=Aut(\langle x\rangle)=\langle \tau\rangle,\ \tau:x\mapsto x^{-1}$$, $$|Aut(\langle x\rangle)|=2$$.
• If $$H\cong \mathbb{Z}_4=\langle y\rangle$$ then we have $$u:\langle y\rangle \to \langle \tau\rangle$$

If $$u$$ is trivial then $$u(y)(x)=x$$ hence $$yxy^{-1}=u(y)(x)=x$$ so $$G\cong \mathbb{Z}_3\times \mathbb{Z}_4$$

If $$u(y)(x)=x^{-1}$$ then $$yxy^{-1}=x^{-1}$$ so $$G=\langle x,y| \ x^3=y^4=1,\ yxy^{-1}=x^{-1} \rangle$$

-If $$H\cong \mathbb{Z}_2\times \mathbb{Z}_2=\langle a\rangle \times \langle b\rangle$$ then we have $$u:\langle a\rangle \times \langle b\rangle\to \langle \tau\rangle$$

If $$u$$ is trivial then $$G\cong \mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_2$$

If $$u(a)(x)=x^{-1}$$ then $$G=\langle a,b,x| \ a^2=b^2=1,\ axa^{-1}=x^{-1},\ bx=xb\rangle$$

Hence we have $$5$$ non isomorphic groups of order $$12$$

Question 1) Is the above proof correct?

Question 2) I know I should have find $$D\cong D_6$$ somewhere but maybe I did something wrong or I can't see the prosentations correctly.

• I think you'll find the question of classifying groups of order twelve has been asked and answered on this site before, maybe even several times. A search would be in order. Aug 22 '20 at 12:21
• E.g., math.stackexchange.com/questions/14754/group-of-order-12 and math.stackexchange.com/questions/1583743/… and various links there. Aug 22 '20 at 12:22
• I reckon your last group is the dihedral group. Aug 22 '20 at 12:23
• @rain1 Maybe I'm wrong (I haven't written it down) but for $|H|=12, H\leq S_4$ it would be the case that $H$ contains all of $3-$ cycles hence $A_4$. Is this correct? Aug 22 '20 at 13:14
• That is great, closes that gap.
– user581023
Aug 22 '20 at 14:47

The proof is good. I think it can be structured more clearly and the groups can be explicitly identified. We know that the following groups exist:

• Abelian: $$C_{12}$$, $$C_2 \times C_2 \times C_3$$.
• Non-Abelian: $$A_4$$, $$D_6$$, $$Dic_3$$ (Also known as the metacyclic group of order 12).

Sylow theory tells us that the Sylow 3-subgroups will be $$C_3$$, and the Sylow 2-subgroups will be $$C_4$$ or $$C_2 \times C_2$$. We also learn that:

• $$n_2 = 1$$ or $$3$$.
• $$n_3 = 1$$ or $$4$$.

When $$n_2 = n_3 = 1$$ we have the Abelian groups.

When $$n_3 = 4$$ you showed that we have $$A_4$$.

We can now look at the only remaining case: $$n_3 = 1$$ and $$n_2 = 4$$. In this situation we are searching for the nontrivial semidirect products $$C_3 \rtimes_\theta P_2$$ with $$\theta : P_2 \to \operatorname{Aut}(C_3)$$.

Note that $$\operatorname{Aut}(C_3) \simeq \langle id, inv \rangle \simeq C_2$$

Let's split into cases based on what $$P_2$$ is.

(Case A) $$P_2 = C_4$$:

In this case there is exactly one nontrivial homomorphism which is forced from $$\theta(0) = 0$$ and $$\theta(1) = 1$$. This gives us the metacyclic group, $$Dic_3$$.

(Case B) $$P_2 = C_2 \times C_2$$:

In this case there are 3 different nontrivial homomorphisms:

• $$\theta_a(0,0) = 0$$, $$\theta_b(0,0) = 0$$, $$\theta_c(0,0) = 0$$
• $$\theta_a(0,1) = 1$$, $$\theta_b(0,1) = 0$$, $$\theta_c(0,1) = 1$$
• $$\theta_a(1,0) = 0$$, $$\theta_b(1,0) = 1$$, $$\theta_c(1,0) = 1$$
• $$\theta_a(1,1) = 1$$, $$\theta_b(1,1) = 1$$, $$\theta_c(1,1) = 0$$

Now these's actually all give us isomorphic semidirect products because we have automorphisms of $$P_2$$ that relate these maps to each other:

• $$(a,b) \mapsto (a,b)$$
• $$(a,b) \mapsto (b,a)$$
• $$(a,b) \mapsto (a,ab)$$

Now we can use $$\theta_a$$ and the definition of multiplication of elements in a semidirect product to see that $$C_3 \rtimes_{\theta_a} C_2 \times C_2 \simeq S_3 \times C_2 \simeq D_{6}$$.