2
$\begingroup$

I'm trying to show that a simple group $G$ is isomorphic to $A_5$.

now I proved that if $G$ has an index 5 subgroup $H$ that is, $|H|=12$, then $G\cong A_5$ using the group action. so I need to know that if $G$ actually has an index 5 group or not.

to show that, I'm actually following this article and it says that if $Q$ is a Sylow 3-subgroup of $G$, then $|N_G(Q)|=15$ for $n_3=4$, the number of Sylow 3-subgroup. and I don't understand why.

since $N_G(Q)$ is a subgroup of $G$, its order should divide $60$ and since any subgroup of $G$ can't have index less than $5$, it means any subgroup of $G$ should have order less than or equal to $12$. and since $|Q|=3$, $3$ divides $|N_G(Q)|$. so we have $|N_G(Q)|=3, 6,12$. I don't know where the number $15$ came out, and why all these $3,6,12$ are removed.

and my second question is that if $A,B$ are two different Sylow $p$-subgroup of any group $D$, then is it true that their normalizer $N_D(A), N_D(B)$ doesn't intersect nontrivially? I felt the author of the article is assuming this.

|cite|improve this question
$\endgroup$
  • $\begingroup$ For the second question, if by 'doesn't intersect nontrivially' you mean 'intersect trivially', this is certainly false in general. Indeed, $N_D(A)$ contains $A$ and $N_D(B)$ contains $B$ and two sylow-p-subgroups need not to be disjoint (so their intersection would contain $A\cap B$) $\endgroup$ – Jef L Jun 3 '17 at 7:59
  • $\begingroup$ @JefLaga what if $A$ and $B$ are of prime order, so they just intersect trivially? can the normalizers of them still intersect nontrivially? $\endgroup$ – user159234 Jun 3 '17 at 8:02
  • $\begingroup$ That might be more subtle, I have to think about that $\endgroup$ – Jef L Jun 3 '17 at 8:28
  • $\begingroup$ Note that in $A_5$ $(1,2)(3,4)$ normalizes both of the Sylow $3$-subgroups $\langle (1,2,5) \rangle$ and $\langle (3,4,5)\rangle$, so normalizers can intersect nontrivially. $\endgroup$ – Derek Holt Jun 3 '17 at 9:32
0
$\begingroup$

All Sylow $3$-subgroups are conjugate. Their number is $n_3$ where $n_3=|G:N_G(Q)|$ where $Q$ is a Sylow $3$-subgroup. So if $n_3=4$ then $|N_G(Q)|=|G|/n_3=4$.

In any case $G$ acts transitively by conjugation on the set of Sylow $3$-subgroups. If $n_3=4$ then $G$ acts on a set of four objects transitively, and so there is a homomorphism $G\to S_4$ whose image is nontrivial. Its kernel is a nontrivial subgroup of $G$, contradiction.

By Sylow's third theorem, $n_3\equiv1\pmod 3$. As $n_3\mid 20$ and $n_3\ge5$ then $n_3=10$. This means that $G$ has $20$ elements of order $3$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ could you explain more about $n_3=|G:N_G(Q)|$? $\endgroup$ – user159234 Jun 3 '17 at 8:08
  • $\begingroup$ $N_G(Q)$ is the stabiliser of $G$ in the conjugation action of $N$ on its Sylow $3$-subgroups. So, it's the orbit-stabiliser theorem. $\endgroup$ – Lord Shark the Unknown Jun 3 '17 at 8:10

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.