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I've been studying for my final exams, and I came across the following question:

If possible, give an example of a simple group $G$ with $n>1$ Sylow $p$-subgroups such that the order of $G$ does not divide $n!$. If not possible, briefly explain why.

Now, it didn't take me long to think of the following "numerical" example: a group of order $60=2^2\cdot3\cdot5$ where there are $3$ Sylow $2$-subgroups. My question is whether or not such a group exists, and if it does, is there any simple way to describe these groups based on knowing how the Sylow structure is built?

In trying to describe the above group, I think I showed that it can't exist. This is because there either have to be $4$ Sylow 3-subgroups or $10$ Sylow 3-subgroups. The second case leads to a contradiction since then there would have to be $k$ Sylow 5-subgroups such that $k(5-1)=30$, and likewise, in the first case, we find we'd have to have 13 Sylow 5-subgroups, also a contradiction. Is all of this correct and is there an easy way to determine whether such a group exists or not?

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    Is ther ...there are $2$ Sylow $2$-subgroups....?? :-)2012-12-15
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    @BabakSorouh: I'm not sure I understand..?? Maybe I'm a bit slow today :)2012-12-15
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    @Clayton: you are probably right - doesn't exist. As to your reasoning, if I did this right, $A_5$ has 10 Sylow-3 groups, so the argument you give for the contradiction (which I didn't quite get) can't be right.2012-12-15
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    Of form $H = (e, (123), (132) )$, of which there are $\binom {5} {3}$.2012-12-15
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    @gnometorule: But $A_5$ has 5 Sylow $2$-subgroups, and $60$ divides $5!$. I specifically wanted $3$ Sylow 2-subgroups, that is where my contradiction comes.2012-12-15
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    You argue below that you cannot have 10 Sylow-3 groups in a simple group of order 60 - or that is what I read your argument to say. That is what I am referring to. $A_5$ does.2012-12-15
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    It can't have $10$ Sylow $3$-subgroups if it has specifically $3$ Sylow 2-subgroups; that is what my argument is geared toward: the construction of a group of order $60$ with $3$ Sylow 2-subgroups.2012-12-15
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    Also, a group of order 2 is no Sylow sub group of $A_5$. The Sylow sub groups have order 3, 4. 5 (with $t$ being 10, 15, 6, so not providing a contradiction unfortunately).2012-12-15
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    Exactly; in the above example, we would have $3\cdot(4-1)=9$ non-identity elements contained in the Sylow $2$-subgroups. Now we have two possible cases for the Sylow $3$-subgroups, of which I explore in the second paragraph above.2012-12-15
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    Sub groups of order 2 are not Sylow; so how can you use that in your argument, specifically referring to the Sylow Theorems?2012-12-15
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    Each Sylow $2$-subgroup has $4$ elements, thus it has $4-1$ non-identity elements. Since I want there to be exactly $3$ Sylow $2$-subgroups, there are $9$ non-identity elements contained in the $3$ Sylow $2$-subgroups.2012-12-15
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    (1) terminology to avoid confusion: what you call a Sylow 2 group is (generally) called a Sylow 4-group. (2) assuming your argument holds. Then no such group of order 60 exists (with $t=4$ for Sylow 4). But $A_5$ is a simple group of order 60 which satisfies what you mean to contradict, so it's no general contradiction to their not being *any* such simple group of order 60. You would need to construct (as you ask early on) an explicit group having such a property; which will probably not work. Stopping to comment now as mathexchange asks me to please do, for a while.2012-12-15
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    Sorry about the confusion; I was unaware. I knew it as a Sylow $p$-subgroup where $p$ was $2$, so naturally I substituted $2$ in for $p$. My apologies.2012-12-15

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The answer is no.

Let $C$ be the set of sylow $p$ subgroups of $G$. It is easy to verify that the map $G\times C\rightarrow C$ that sends $(x,H)$ to $xHx^{-1}$ is an action of $G$ on $C$. By the orbit stabilizer theorem, we know that: $$\forall H\in C[|C||N(H)|=|G|]$$ Since $1<|C|$, we deduce that $|N(H)|<|G|$ for all $H\in C$....(1)

By sylow's theorem we know that the function $T_x:C\rightarrow C$ that sends $H$ to $xHx^{-1}$ is a permutation of $C$. Now consider the homomorphism $\phi:G\rightarrow S_C$ that sends $x$ to $T_x$.

Claim: $\phi$ is an injection.

Proof: We know that $\operatorname{Im}\phi\cong G/\ker\phi$. Thus, it suffices to show that $|\ker\phi|=1$. This is easy to see because $\ker\phi\unlhd G$, since $G$ is simple, therefore either $\ker\phi=\{e\}$ or $\ker\phi=G$. Let $H\in C$, it is easy to verify that $\ker\phi \leq N(H)$. Thus, if $\ker\phi=G$, we would conclude that $N(H)=G$. However, we know from (1) that $N(H)$ is a proper subgroup of $G$. Therefore, we finally deduce that $\ker\phi=\{e\}$.

Thus, $G$ is isomorphic to a subgroup of $S_C$ (The group of permutations of $C$). Hence, $|G|\,\Big|\,|S_C|!=n!$

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    Thank you! Just to check, your line $|C|\cdot|N(H)|=|G|$ follows from the fact that the number of conjugate subgroups of $H$ is equal to the index of the normalizer?2012-12-17
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    Yes. Which also follows from the orbit stabilizer theorem2012-12-17