Prove that the group of permutations of four symbols $S_4$ contains a normal subgroup H such that the quotient group $S_4/H$ is isomorphic to the group of permutations of three symbols $S_3$.
$S_4$ has order $24$. Any subgroup thus will have order $1, 2, 3, 4, 6, 8, 12$ or $24$.
A normal subgroup is a union of conjugacy classes which in this case correspond to the cycle shapes as follows:
- 6 of the form $(abcd)$
- 8 of the form $(abc)(d)$
- 3 of the form $(ab)(cd)$
- 6 of the form $(ab)(c)(d)$
- 1 identity element
If $S_4/H\simeq S_3$ with one of the above normal subgroups as $H$ then by Lagrange $|H|=\frac{|S_4|}{|S_3|}=4$ but there are no normal subgroups with this order.
Would you be able to help me with this?