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What are the normal subgroups of $G=S_4\times S_3$.

I think $G$ has a normal subgroup of order $72$ is there any other nontrivial normal subgroups of $G$.

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    $\begingroup$ A Cartesian product $A\times B$ of two groups $A$ and $B$ has as normal subgroups $A\times\{e_B\}$ and $\{e_A\}\times B$ where $e_A, e_B$ are the identity elements. In some cases there are also other normal subgroups. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 11 '14 at 1:56
  • $\begingroup$ @Michal Hardy..is that 2-Sylow or 3-Sylow subgroups of G are normal subgroups of this group..thanks for your time $\endgroup$ – suresh p Dec 11 '14 at 2:04
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    $\begingroup$ Neither the Sylow $2$-subgroups nor the Sylow $3$-subgroups are normal subgroups. It is's any help, there are $13$ normal subgroups altogether, and their orders are $ 1, 3, 4, 6, 12, 12, 24, 24, 36, 72, 72, 72, 144$, so there are three of order $72$. $\endgroup$ – Derek Holt Dec 11 '14 at 9:16
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If we have $N\lhd G$ and $M\lhd H$. Then we have $N\times M \lhd G\times H$.

Take $A_4\times S_3$

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  • $\begingroup$ Hi..i need prove for order of 72 is normal subgroup of G. can u help me $\endgroup$ – suresh p Dec 11 '14 at 2:17
  • $\begingroup$ Take the direct product $A_4\times S_3$, this has order $12\cdot 6$ which is $72$, notice this is normal since we have $A_4\lhd S_4$ and $S_3\lhd S_3$ $\endgroup$ – HereToRelax Dec 11 '14 at 2:21
  • $\begingroup$ why the downvote? $\endgroup$ – HereToRelax Dec 11 '14 at 4:30
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    $\begingroup$ The downvote is presumably because what you wrote is not true. For example $C_2 \times C_2$ has a normal subgroup that is not of this form. It's not even true in the example in the question. $\endgroup$ – Derek Holt Dec 11 '14 at 9:12
  • $\begingroup$ Oh. My bad. I think it is ok now, do you agree? $\endgroup$ – HereToRelax Dec 11 '14 at 18:25
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Let me give you a hint. Suppose we are looking for normal subgroups of $A\times B$. It is reasonable that we should expect this to relate in some way to the normal subgroups of $A$ and the normal subgroups of $B$. If $X$ is a subgroup of $A$, there's a subgroup of the form $X\times 1$ in $A\times B$. Let's conjugate $X\times 1$ by a generic $(a,b)\in A\times B$. $$(x,1)\in X\times 1 \hspace{10pt}\Rightarrow \hspace{10pt} (a^{-1},b^{-1})(x,1)(a,b)=(a^{-1}xa,b^{-1}1b)=(a^{-1}xa,1)$$

  • When is it true that this will be an element of $X\times 1$?

  • What if we apply this same logic to $X\times Y$, where $X\subset A$ and $Y\subset B$?

  • Can you apply this now to $S_4\times S_3$?

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