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.

  • 1
    $\begingroup$ 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. $\endgroup$ Aug 22, 2020 at 12:21
  • $\begingroup$ E.g., math.stackexchange.com/questions/14754/group-of-order-12 and math.stackexchange.com/questions/1583743/… and various links there. $\endgroup$ Aug 22, 2020 at 12:22
  • $\begingroup$ I reckon your last group is the dihedral group. $\endgroup$ Aug 22, 2020 at 12:23
  • 1
    $\begingroup$ @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? $\endgroup$
    – 1123581321
    Aug 22, 2020 at 13:14
  • 1
    $\begingroup$ That is great, closes that gap. $\endgroup$
    – user581023
    Aug 22, 2020 at 14:47

1 Answer 1


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: $A_4$, $D_6$, $\mathrm{Dic}_3$ (Also known as the dicyclic, or 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 $C_3 \rtimes_{\theta_a} C_2 \times C_2 \simeq S_3 \times C_2 \simeq D_{6}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .