Reference for a Proposition: "if a subgroup $H$ contains a normal $p$-Sylow $ P$, then $H \subset N_G(P)$, a normalizer of a $p$-Sylow in $G$"

Question

Let's first introduce the reference I'm trying to reformulate the proposition from:

How to enumerate subgroups of each order of $S_4$ by hand

For example , for a subgroup of order 6 , the writer explains :

A subgroup of order 6 must have a normal Sylow 3-subgroup, so must live inside the normalizer (inside S4) of a Sylow 3-subgroup. The Sylow 3-subgroups are just the various alternating groups of degree 3, and their normalizers are various symmetric groups of degree 3, so are exactly the 4 subgroups of order 6.

I can't really figure out where is this result coming from , what proposition , theorem it refers to , or just logical argument. If I try to reformulate what I understand in the specific case of $S_4$:

If a subgroup $H$ of $S_4$ contains a normal $p$-Sylow $P$, then $ H\subset N_{S_4}(\rho)$, with $\rho$ a $p$-Sylow of $S_4$ and $N_{S_4}(\rho)$ the normalizer of one of these $p$-Sylow in $S^4$.

I'd like either a correction or a reference for this proposition.

I may be formulating it really bad or maybe I don't get the point in this quote, however, the title of the question will be modified when solved for future readers to be able to refer to it if needed.

Answer 1

I am reading that passage from Jack Schmidt's answer as follows.

It calls upon the following

Fact. If $G$ is a group, $H\le G$ a subgroup, and $K\unlhd H$ yet another subgroup of $G$ that also happens to be a normal subgroup of $H$, then $H\le N_G(K)$. In other words, $H$ is contained in the normalizer of $K$ in $G$.

This hardly needs a proof. The assumption $K\unlhd H$ is equivalent to $hK=Kh$ for all $h\in H$, but the normalizer is $$ N_G(K)=\{x\in G\mid xK=Kx\} $$ by definition, so $H\le N_G(K)$.

Anyway, I fill in extra details to the steps of that passage.

• A subgroup of order $6$ must have a normal Sylow $3$-subgroup, This can be seen using general Sylow theory, but that feels a bit like swatting flies with cannonballs. Your first course on algebraic structures may have had the result that a group of order six is either cyclic, $C_6=\langle c\rangle$, or the symmetric group $S_3$. Both have a normal Sylow $3$-subgroup. Namely $\langle c^2\rangle$ and $A_3=\langle(123)\rangle$ respectively.
• so must live inside the normalizer of a Sylow $3$-subgroup (of $S_4$). Here the Fact is applied. If $H\le S_4$ is an arbitrary subgroup of order six then by the previous bullet it has a normal Sylow $3$-subgroup, call it $P$. In particular, $|P|=3$. Because $3^2\nmid 24=|S_4|$, it follows that $P$ is also a Sylow $3$-subgroup of $G=S_4$. As $P\unlhd H$, the Fact tells us that $H\le N_{S_4}(P)$.
• The Sylow 3-subgroups are just the various alternating groups of degree 3, Sylow's theorems tell us that $S_4$ can have either $1$ or $4$ Sylow $3$-subgroups. It is easy to exhibit four: $\langle(123)\rangle$, $\langle(124)\rangle$, $\langle(134)\rangle$ and $\langle(234)\rangle$, each stabilizing one of the four numbers. Therefore the number of Sylow $3$-subgroups of $S_4$ is exactly four.
• and their normalizers are various symmetric groups of degree 3, Because the Sylow subgroups are all conjugate (in $S_4$), the orbit-stabilizer theorem tells us that their normalizers all have order $24/4=6$. Obviously each of the listed four Sylow subgroups is normalized by its respective copy of $S_3$. For example $N_{S_4}(\langle(134))\supseteq Sym\{1,3,4\})$. Because the normalizer has order six as does the symmetric group, we must have equality here.
• so are exactly the 4 subgroups of order 6. We have seen that any subgroup $H\le S_4$ of order six normalizes a Sylow $3$. We have also seen that the normalizer of any Sylow $3$ has exactly six elements. Therefore we can conclude that $S_4$ has exactly four subgroups of order six, and we have accounted for all of them.