Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What are the normal subgroups of order $12$ in $S_{3} \times S_{3}$?

I know that all the subgroups of order $12$ in $S_{3} \times S_{3}$ are isomorphic to the dihedral group of order $12$.

share|cite|improve this question
up vote 3 down vote accepted

The derived group of a direct product $G \times H$ is $[G,G] \times [H,H].$ The derived group of $S_{3}$ is $A_{3}.$ Hence the derived group of $X = S_{3} \times S_{3}$ has index $4.$ Every normal subgroup $Y$ of $X$ such that $X/Y$ is Abelian contains $[X,X],$ so has index $1,2$ or $4.$ Hence $X$ has no normal subgroup of index $3,$ ie of order $12.$

share|cite|improve this answer
Thank you very much. – user28083 Sep 17 '12 at 8:26

Normal subgroups are closed under intersection, so any normal subgroup of $S_n \times S_n$ must restrict to a normal subgroup on each coordinate. For $n = 3$ or $n \geq 5$, the only nontrivial proper normal subgroup of $S_n$ is the alternating group $A_n$. This allows you to enumerate the possible orders of normal subgroups of $S_n \times S_n$.

share|cite|improve this answer
Thank you very much. – user28083 Sep 17 '12 at 8:31

Suppose $S_3=\mathrm{Sym}\{a,b,c\}$, i.e. $S_3=\{id,(ab),(bc),(ca),(abc),(acb)\}$. Then the transpositions are each other's conjugate, as well as the rotations.

If $N\le S_3\times S_3$ is normal, it is closed on conjugation, so, if for example $\langle (ab),(bc)\rangle\in N$, then first conjugating in the first coordinate (by appropriate $\langle g,id\rangle$), then in the second coordinate, we obtain that $N$ must contain all pairs of transpositions, which generate the whole $S_3\times S_3$.

I would then look for the normal subgroup generated by an element like $\langle (ab),(abc)\rangle$ , and so on..

.. For larger groups, you may use some Sylow theorems..

share|cite|improve this answer
Thank you very much. – user28083 Sep 17 '12 at 8:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.