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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.

Arlon Fredolster
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    This is true for any normal subgroup, not just normal Sylow $p$-subgroups. If $N$ is normal in $H$, then $H$ normalizes $N$, hence $H$ is contained in the normalizer of $N$. –  Dec 28 '20 at 11:35
  • This notation is kind of weird: $N_{\rho}(S_4)$. Do you mean $N_G(Q)$ where $Q$ is some Sylow subgroup of $G$? – Nicky Hekster Dec 28 '20 at 11:52
  • Yes sorry , i'm correcting it. – Arlon Fredolster Dec 28 '20 at 13:17
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    The trouble with the original proposition and with your reformulation is this: it is not clear to which group "is a Sylow subgroup" applies, is it Sylow in the small group or Sylow in the big group. – ancient mathematician Dec 28 '20 at 13:44
  • Yes ancientmathematician , that's what's confusing me a lot , and why i'm not sure to understand the original idea in this quote.. – Arlon Fredolster Dec 28 '20 at 15:06
  • I think your problem is that you could not read which group is a Sylow sub of which, and which is contained in the normalizer of some other in the given bigger group. In my opinion the context allows for only a single interpretation. I tried to break it into smaller pieces in my answer. I'll be teaching this stuff in a few weeks time, so writing down the details is somewhere between a warm-up and a penance (for not being busily updating my lecture notes already). – Jyrki Lahtonen Dec 30 '20 at 23:09

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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.
Jyrki Lahtonen
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