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Find all normal subgroups of $S_3 \times S_3$.

What are normal subgroup and $S_3 \times S_3$? Could I have some examples?

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Ah, a classic. This question is tricky: aside from the obvious subgroups that are the products of subgroups of $S^3$, there is a hidden one. But revealing much more will ruin the fun! – user18063 Jan 7 '12 at 22:21
"What are normal subgroup and $S_3 \times S_3$?" -- This is quite unclear. Do you want to know what "normal subgroup" means, or what the group $S_3 \times S_3$ looks like? If yes, then it might be better to go back and learn these concepts, and/or ask a question specifically about them. – Srivatsan Jan 7 '12 at 23:11
The question in the subject line seems perfectly clear, but it looks as if maybe the poster doesn't understand it. – Michael Hardy Jan 7 '12 at 23:30
up vote 8 down vote accepted

Just like in any group there are two "obvious" normal subgroups (the trivial subgroup and the whole group), but there may be others, in a direct product there are some "obvious" normal subgroups, but there may be others.

If $G$ and $H$ are groups, and $N\triangleleft G$ is any normal subgroup of $G$ and $M\triangleleft H$ is any normal subgroup of $H$, then $N\times M$ is a subgroup of $G\times H$, and in fact $(N\times M)\triangleleft (G\times H)$. I don't know which of the many equivalent definitions of normal subgroup you are using, nor what theorems you may have already proven about normal subgroups, so I'll leave the verification to you.

So, for example, since the normal subgroups of $S_3$ are $\{1\}$, $A_3$, and $S_3$, then you automatically get nine normal subgroups of $S_3\times S_3$ by taking all subgroups: $\{1\}\times\{1\}$ (the trivial subgroup), $\{1\}\times A_3$, $\{1\}\times S_3$, $A_3\times\{1\}$, $A_3\times A_3$, $A_3\times S_3$, $S_3\times\{1\}$, $S_3\times A_3$, and $S_3\times S_3$ (the whole group).

But there may be others (just like there are subgroups of $G\times H$ that are not of the form $A\times B$ with $A\leq G$ and $B\leq H$). To give you a different example, take $C_2\times C_2$, there $C_2$ is the cyclic group of order $2$. The only subgroups of $C_2$ are $\{1\}$ and $C_2$, so you only get four normal subgroups of $C_2\times C_2$ this way: $\{1\}\times\{1\}$, $\{1\}\times C_2$, $C_2\times\{1\}$, and $C_2\times C_2$. But there is another (normal) subgroup of $C_2\times C_2$: $\{ (1,1), (x,x)\}$; this does not come as a product of two normal subgroups.

So you still need to check to see if, in addition to the nine normal subgroups of $S_3\times S_3$ that we found above, there may be another.

To get you started, assume that $N\triangleleft S_3\times S_3$ is not the trivial subgroup, and $(a,b)\in N$. If $a\neq 1$, consider conjugation by elements of the form $(x,1)$ to try to figure out what is the collection of all elements of the form $(y,1)$ that are in $N$. Then work symmetrically to find elements of the form $(1,z)$.

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Finding some normal subgroups is not hard -- in addition to the ones Arturo mentioned there is (at least) also the group consisting of $(a,b)$ where the permutations $a$ and $b$ have the same sign. However, groping about for examples is not going to help you prove that you have all normal subgroups.

For that, it seems feasible in this case to remember that a normal subgroup is a union of conjugacy classes. Since $S_3$ has only 3 conjugacy classes (the identity, the transpositions, and the 3-cycles), there are 9 conjugacy classes in $S_3\times S_3$, and with a bit of effort you can systematically check which combinations of them are subgroups. You can reduce the number of cases to check significantly by noting that $\{(1,1)\}$ must always be included and that the sum of the sizes of the conjugacy classes you include must divide 36.

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