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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.

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  • $\begingroup$ 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
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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 $3$.

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  • $\begingroup$ 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

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