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Let $G=S_4\times S_3$. Then

(1) a 2-sylow subgroup of $G$ is normal

(2) a 3-sylow subgroup of $G$ is normal

(3) $G$ has a non trivial normal subgroup

(4) $G$ has a normal subgroup of order 72

I tried to apply sylows theorems for $G$. $|G|=4\cdot 3^2\cdot 2^2$. Then $G$ has 2-sylow subgroup and 3-sylow subgroup of order $2^2$,$3^2$. I used sylows second theorem to see number of those sylows subgroups. But i can not conclude. Please help me some one.

S. Pitchai Murugan
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  • Hint: both $;S_4\times 1;,;;1\times S_3;$ are normal in $;G;$ , and any normal subgroup of each of them will be normal in the direct product. For example, $;1\times A_3\lhd G;$ . – Timbuc Oct 16 '14 at 12:08
  • Another hint: Can you show that the set $$K={(\alpha,\beta)\in S_4\times S_3\mid sgn(\alpha)\cdot sgn(\beta)=1}$$ is a normal subgroup of $S_4\times S_3$? Here $sgn$ is the sign of a permutation. – Jyrki Lahtonen Oct 17 '14 at 13:04

1 Answers1

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First of all the 2-Sylow subgroups of G are of order $16$ and not $4$. Now in order to construct a Sylow-2 subgroup $T_2$ of $G$ we first note that $T_2 \cap S_4$ is a Sylow-2 subgoup of $S_4$ with order $8$. The only subgroups of order $8$ are the dihedral-8 subgroups but these are not normal in $S_4$ since there are $3$ conjugates. In order to find the whole of $T_2$ we have to consider $T_2 \cap S_3$ which is necesseraly $C_2$. We conclude that $T2$ can be constructed by taking the direct product of a dihedral subgroup of $S_4$ with one of the three cyclic-2 subgoups of $S_3$ to obtain nine conjugate subgroups of $T2$ so that $T2$ is not normal in $G$. A similar construction of $T3$ shows that it is the direct product of one of the four cyclic-3 subgroups of $S_4$ with the unique subgroup of $S_3$ so that $T_3$ isn't normal either. To prove that a normal subgroup of $G$ with order $72$ exists we remark that $A_4$ is normal in $S_4$ so that $A_4 \times S_3$ is normal in $G$.

Marc Bogaerts
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  • Why is the intersection of T2 with S4 a 'sylow' subgroup(of S4)? I know that its order will divide 8 but why is it exactly 8? – jimm Nov 07 '20 at 06:15