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Show the group $ \ S_4 \times S_3 \ $ has a normal subgroup of order $ \ 72 \ $.

Answer:

Since $ A_4 \ $ is a normal subgroup of $ S_4 \ $ , $ \ \ \ A_4 \times S_3 \ $ is a subgroup of $ S_4 \times S_3 \ $.

Now $ o(A_4 \times S_3 ) \ =12 \times 6=72 $ , which is the half of the order of $ S_4 \times S_3 \ $.

Hence $ A_4 \times S_3 \ $ is a normal subgroup of $ S_4 \times S_3 \ $ of order $ 72 \ $ .

My question is - why did not we consider here the subgroup $ S_4 \times A_3 \ $ whose order also $ 72 \ $ ?

Is $ \ S_4 \times A_3 \ $ a subgroup of $ \ S_4 \times S_3 \ $ ?

If not , then how would I show this ?

I am confused.

Is there any help ?

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  • $\begingroup$ We could've looked at $S_4\times A_3$ just as well, it is a normal subgroup of order $72$. The question is not "find all normal subgroups of order $72$", it is "show that there is at least one". $\endgroup$ – Wojowu Sep 30 '17 at 14:40
  • $\begingroup$ I know $ A_4 \times S_3 $ is normal subgroup of order 72 . . But My question is -why did not consider the subgroup $ S_4 \times A_3 \ $ ? $\endgroup$ – M. A. SARKAR Sep 30 '17 at 14:45
  • $\begingroup$ We could've considered either. It doesn't matter. $\endgroup$ – Wojowu Sep 30 '17 at 14:46
  • $\begingroup$ There's no reason why not consider $S_4\times A_3$ though... $\endgroup$ – user441558 Sep 30 '17 at 14:47
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    $\begingroup$ Of course $S_4 \times A_3$ is a normal subgroup. It has index $2$. $\endgroup$ – Bungo Oct 1 '17 at 2:42

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