# Normal subgroups of order 12 in $S_3\times S_3$

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 $12$.

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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 $12.$

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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 $S_n \times S_n$.

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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..

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Thank you very much. – user28083 Sep 17 '12 at 8:27