How to prove that $Q_8$ is not subgroup of $S_4$ by using Sylow theorem?

Question

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 $S_4$?

By Sylow's theorem, all Sylow 2-subgroups are conjugate, so they are isomorphic to $D_4$. — lEm, May 17 '19 at 05:39
Ah ha! Let the $Q_8$ is a subgroup of $S_4$, then $Q_8$ is a sylow 2 group. But since $Q_8$ and $S_4$ are not isomorphic, it's contradiction. Is my guess right? — se-hyuck yang, May 17 '19 at 07:33

score 3 · Accepted Answer · edited Sep 25 '23 at 08:32

3

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 $n_2=3$.

edited Sep 25 '23 at 08:32

user26857

52,094

answered May 17 '19 at 05:53

user10354138

33,239

You meant there are some elements in $S_4$(not in $D_4$) Whose order is 4. So $n_2$ is 3. Am I right? – se-hyuck yang May 17 '19 at 07:48

How to prove that $Q_8$ is not subgroup of $S_4$ by using Sylow theorem?

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