# Product of two symmetry groups

Do there exist 2-sylow subgroups of $S_4\times S_3$ that are normal?

Do there exist 3-sylow subgroups of $S_4\times S_3$ that are normal?

Thank you for helping!

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Fact $1$: Suppose $G$ and $H$ are finite groups and $P_G$ and $P_H$ are Sylow $p$-subgroups of $G$ and $H$, respectively. Then $P_G \times P_H$ is a $p$-Sylow subgroup of $G \times H$.

Fact $2$: $A \times B \trianglelefteq G \times H$ if and only if $A \trianglelefteq G$ and $B \trianglelefteq H$.

Fact $3$: If there is at least one Sylow $p$-subgroup that is not normal in $G$, then $G$ has no normal Sylow $p$-subgroup.

Fact $4$: $S_4$ does not have a normal Sylow $3$-subgroup and $S_3$ does not have a normal Sylow $2$-subgroup.

From this you can conclude that $S_4 \times S_3$ does not have a normal Sylow $2$-subgroup or a normal Sylow $3$-subgroup.

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$$H_k:=\langle\,\left((12k),(1)\right)\,,\,\left((1),(123)\right)\,\rangle \leq S_4\times S_3\,\,,\,\,k=3,4$$ are two different Sylow 3-subgroups (order 9) of $\,S_4\times S_3\,$ and, thus, there is not such one normal.

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No. The list of normal subgroups can easily be calculated, for instance using GAP. Notice none of the normal subgroups has the correct order to be a Sylow subgroup.

gap> K := SymmetricGroup(3);
Sym( [ 1 .. 3 ] )
gap> H := SymmetricGroup(4);
Sym( [ 1 .. 4 ] )
gap> G := DirectProduct(H,K);
Group([ (1,2,3,4), (1,2), (5,6,7), (5,6) ])
gap> List( NormalSubgroups(G), Size );
[ 144, 72, 72, 72, 24, 36, 24, 12, 12, 6, 3, 4, 1 ]

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I used a List() command to shrink the code, and added a little explanation to what the calculation shows. @Tim Duff: Feel free to revert if you don't like it. – Jack Schmidt Jul 7 '12 at 13:48
Fancy - I like it. Guess it's time to learn more GAP... – user17794 Jul 8 '12 at 0:00