Prove that $Q_8$ is not subgroup of $S_4$.
It is trivial that $D_4$ is a subgroup of $S_4$.
But the case $Q_8$ make me confuse why this group is not a subgroup of the $S_4$.
Why $Q_8$ isn't a subgroup of $S_4$ ?
Many people might proved those statement either Group action or using mapping.
But I want to know prove those statement using Sylow theorem.
PS How many $P_2$ (Sylow 2-subgroup of $S_4$ whose order is 8) are in $S_4$?
Best Answer
By $D_n$ I assume you mean the dihedral group of order $2n$.
Note that a Sylow-2 of $S_4$ has order $8$, so we know $D_4$ is a Sylow-2 of $S_4$. Since any two Sylow-$p$ are conjugate, they are isomorphic. Thus all subgroups of order $8$ in $S_4$ have to be isomorphic to $D_4$.
Note that $n_2\equiv 1\pmod{2}$, $n_2\mid \#S_4$, so either $n_2=1$ or $n_2=3$. But the number of 4-cycles in $S_4$ is 6, and an $D_4$ only has two element of order 4. Since any group of order 4 appears as a subgroup of a Sylow-2, we cannot have $n_2=1$ and hence $n_2=3$.