3
$\begingroup$

I am reading a proof to the problem stated above and I have a few questions that I hope somebody helps me clarify. THe proof goes as follows:

Let $G$ be a subgroup of order 224 and $S$ be the set of $7$ Sylow subgroups of $G$. THe cardinality of $S$ divides 32 and is congruent to 1 mod 7, so has to be 1 or 8. Let $H$ be a 2-Sylow subgroup of $G$. It has 32 elements and acts on $S$ by conjugation.

This is my first question, how do we know that $H$ acts on $S$ by conjugation. My guess is that since $S$ is the set of $7$-sylow subgroups and the $7$ sylow subgroups are conjugate, then action by conjugation of an element $h \in H$, will just send one 7 Sylow subgroup to another one and so the conjugation of $S$ will just be $S$. Is this right or is there any other reason to have this action? If it is not right, how could we know when to use the action by conjugation?

Let $N\in S$ be a $7$-Sylow subgroup, and let $U$ be its stabilizaer subgroup in $H$. The $H$-orbit of $N \in S$ has at most 8 elements, so the stabilizer has $32/8=4$ elements.

Here I have another question, why does the stabilizer have just 4 elements?

The group $U$ normalizes $N$. Let $\phi:U \rightarrow Aut(N)$ be the group homomorphism associated with the conjugation action of $U$ on $N$. Since $Aut(N)$ has $6$ elements and the number of elements of $U$ is divisible by $4$, the group homomorphism $\phi:U \rightarrow Aut(N)$ cannot be injective.

It is still not clear for me why $\phi$ cannot be injective. Can someone explain the relationship between $Aut(N)$ having 6 elements, $U$ being divisible by $4$ and $\phi$ not being injective?

We can choose a nontrivial element of order 2 in the kernel of $\phi$. We can also choose $g$ be a generator of $N$. Then $h$ and $g$ commute,$h$ has order 2 and $g$ has order 7. Then $hg$ has order 14.

$\endgroup$
1
$\begingroup$

By Sylow's second theorem, the Sylow $p$-subgroups of $G$ are all conjugate under $G$. Therefore they form one orbit under the action of conjugation. Let $N$ be a given Sylow subgroup. Then, $NH=G$, and $K$ fixes itself under conjugation, so the $H$-orbit of $N$ is the same as the $G$-orbit of $N$ (in the conjugation action) so $H$ acts transitively on the Sylow $7$-subgroups.

If there are $8$ Sylow subgroups, the stabiliser of $N$ in $H$ has order $|H|/8=4$.

If there is an injective homomorphism $\phi:A\to B$ then $|A|\mid|B|$. This is Lagrange's theorem, applied to $\phi(A)$.

ADDED IN EDIT

Here's another way of proving $G$ has an element of order $14$. As in the question, we see $G$ has either $6$ or $48$ elements of order $7$. The subgroup $H$ of order $32$ acts on these elements of order $7$ by conjugation. It's impossible for all the orbits of this action to have order $32$, so act least one has order a smaller power of $2$. So there is $a\in G$ of order $7$ and a non-identity $b\in H$ with $bab^{-1}=2$. Some power $c$ of $b$ has order $2$. Then $cac^{-1}=a$. Then $a$, $c$ commute and have orders $7$ and $2$ respectively. Then $ac$ has order $14$.

$\endgroup$
0
$\begingroup$

1) Any subset $\;X\;$ of $\;G\;$ acts on $\;S\;$ by conjugation since for any $\;x\in X,\,s\in S\,,\,\,x^{-1}sx\in S\;$ . This is true even without Sylow theorems, as conjugation (being an isomorphism) keeps order of subgroups.

2) We (must) know that for any action of a finite group of a finite set, $\;|\mathcal Orb(s)|=[G\,:\,Stab(s)]=\frac{|G|}{|Stab(s)|}\;$ . Since the left hand is either one or eight, the right hand is either one or eight, thus $\;|Stab(s)|\;$ is either $\;32\;$ or $\;4\;$.

3) You have a homomorphism $\;\phi: U\to\text{ Aut}\,N\;$ . If $\;|U|=32\;$ then $\;\phi\;$ clearly can't be injective, as the automorphism group of a group of order a prime is that prime minus one, thus $\;|\text{Aut}\,N|=7-1=6\;$ . Otherwise, $\;|U|=4\;$ , so it can't inject into a group of order 6 by Lagrange's theorem.

$\endgroup$
0
$\begingroup$

Question 1: Yes, that's exactly it. The entire group $G$ acts on $S$ by conjugation, so so does any subgroup of $G$.

Question 2: They do not say "just". What they say is "so the stabilizer has $32/8=4$ elements." There is no "exactly" in there. There is no "just". That means it is to be taken to mean "at least". They are saying that 4 distinct elements exist in that subgroup, but they make no implications about whether those are all the elements. They even say later "the number of elements in $U$ is divisible by $4$" not "is $4$".

Question 3: Lagrange's theorem. The image of $\phi$ is a subgroup of $Aut(N)$, and therefore its order must divide $6$. If $\phi$ were injective then the image of $\phi$ would be isomorphic to $U$ and therefore be divisible by $4$. No number divides $6$ and is divisible by $4$, so $\phi$ cannot be injective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.