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I'm trying to understand this proof, but there are some strange things that I didn't understand.

Why $M:=P_3\cap Q_3$ implies that $|M|$ can't be $9$? As I understand, the $3$-Sylow subgroups for groups of order $90$ must have $9$ elements. Also, what is an intersection of two $p$-Sylow subgroups?

What is $|\langle P_3,Q_3\rangle|$? How did he arrive at the order being $45$ or $90$?

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  • $\begingroup$ Assuming that the group is simple, it must have at least two distinct $3$-Sylow subgroups, that are denoted by $P_3$ and $Q_3$. Their intersection is smaller than both, so it has cardinality $1$ or $3$. $\endgroup$ – egreg Jun 19 '17 at 22:23
  • $\begingroup$ $H = \langle P_3, Q_3 \rangle$ is the subgroup of $G$ generated by $P_3$ and $Q_3$. See the comment below the response. It was shown in the answer that since the subgroup of $H$ contains $P_3Q_3 = \{ab | a \in P_3, b \in Q_3\}$, which by groupprops.subwiki.org/wiki/Product_of_subgroups has order at least $27$ as shown in the response. Since $H$ is a subgroup its order must divide $90$, so it is either $30, 45,$ or $90$. However, $H$ contains $P_3$ and $Q_3$ as subgroups, so its order must be a multiple of $9$, so it cannot be $30$. $\endgroup$ – user357980 Jun 19 '17 at 22:40
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Here are some attempted answers to your specific questions.

1) Why $M=P_3\cap Q_3$ can't have order 9.

This is as egreg said in the comments. If there were a single sylow 3-subgroup, it would be normal, and so the whole group wouldn't be simple. So in a hypothetical simple group of order 90, it must be that there are at least 2 different (not equal to each other) sylow 3-subgroups, call them $P_3$ and $Q_3$, both with 9 elements. Their intersection is a subgroup $M$ contained in both of them. If it had 9 elements, it would have to be equal to $P_3$ or $Q_3$ and then they would have to be equal to each other. So it must have 1 or 3 elements.

2) What is an intersection of two sylow $p$-subgroups?

It's just the set of elements that are in both. (Just like the intersection of any two subgroups. It is necessarily a subgroup. [Why?])

I'm not exactly sure I'm speaking to what's bothering you.

3) What is $|\langle P_3,Q_3\rangle |$?

We may as well start with "what is $\langle P_3,Q_3\rangle$?" This is the subgroup generated by $P_3$ and $Q_3$. I.e. it is the smallest subgroup of $G$ containing both $P_3$ and also $Q_3$.

$|\langle P_3,Q_3\rangle |$ is the order of this subgroup.

4) How did he arrive at the order being 45 or 90?

Here's the key idea. The subgroup generated $P_3$ and $Q_3$ must, at the least, contain all pairwise products $pq$ where $p\in P_3$ and $q\in Q_3$. The set of these pairwise products is called $P_3Q_3$. By a standard counting principle (e.g. see section 2.5 in I. N. Herstein's Topics in Algebra), the number of distinct elements in $P_3Q_3$ is equal to

$$\frac{|P_3||Q_3|}{|P_3\cap Q_3|} = \frac{9\cdot 9}{|M|}$$

which from number (1) above we know must be at least $81/3=27$.

This means $\langle P_3,Q_3\rangle$ (the subgroup generated by $P_3$, $Q_3$) is a subgroup of $G$ with at least 27 elements. Since it is a subgroup, its order is a factor of 90. The only factors of 90 that are at least 27 are 30, 45, and 90.

DonAntonio left it as an exercise for the reader to rule out the possibility of 30. 30 can be ruled out because $\langle P_3,Q_3\rangle$ is a subgroup containing $P_3$ and $Q_3$, which have 9 elements, but 30 is not a multiple of 9, thus a group of order 30 cannot contain a group of order 9.

This leaves 45 and 90.

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