Let $G$ be the group $S_4\times S_3$ .Prove that $G$ has a normal subgroup of order $72$.

Attempt:Order of $G$ is $144=2^4.3^2$.So $G$ has a Sylow $2$ subgroup of order $16$ and a Sylow $3$ subgroup of order $9$.Number of Sylow $2$ subgroups $=1+2k=n_2$ divides $9$ and similarly $n_3$ divides $16$.

So $n_2=1,3,9$ and $n_3=1,4,16$.These arguements are taking me nowhere towards solving this problem.

How to approach this problem?

  • 1
    $\begingroup$ Can you say anything about what normal subgroups in a direct product look like? $\endgroup$ – Max Nov 27 '15 at 16:47

take $S_4\times A_3$ this is a subgroup of order $72.$ Since any subgroup of index 2 is normal, the above subgroup is normal.

  • $\begingroup$ thank you very much ;that helped a lot $\endgroup$ – Learnmore Nov 27 '15 at 17:34
  • $\begingroup$ you are wellcome @learnmore $\endgroup$ – Black-horse Nov 27 '15 at 17:49

Or, take $A_4 \times S_3$, another normal subgroup of order $72$. Note that $A_4 \times S_3 \ncong S_4 \times A_3$ (take the centers, the left hand side has trivial center, the right hand side has a center of order $3$). However the respective quotient groups are isomorphic (to $C_2$).


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