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

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

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 ?

• 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". – Wojowu Sep 30 '17 at 14:40
• 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 \$ ? – M. A. SARKAR Sep 30 '17 at 14:45
• We could've considered either. It doesn't matter. – Wojowu Sep 30 '17 at 14:46
• There's no reason why not consider $S_4\times A_3$ though... – user441558 Sep 30 '17 at 14:47
• Of course $S_4 \times A_3$ is a normal subgroup. It has index $2$. – Bungo Oct 1 '17 at 2:42