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Let $p$ be a prime and $H$ a subgroup of a finite group $G$. Let $P$ be a p-sylow subgroup of G. Prove that there exists $g\in G$ such that $H\cap gPg^{-1}$ is sylow subgroup of $H$.

I have no idea how to do this, any hints?

Note: Originally it was unclear if the problem was for possibly infinite groups or just finite ones. However, since the definition of $p$-Sylow subgroup being used is that it is a $p$-subgroup such that the index and the order are relatively prime, the definition only applies to finite groups.

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$G$ need not be finite. –  Chu Dec 29 '10 at 18:00
    
edit the question and add the information that $G$ need not be finite there. –  Mariano Suárez-Alvarez Dec 29 '10 at 18:35
    
thanks. It's been added. –  Chu Dec 29 '10 at 18:41
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4 Answers

up vote 3 down vote accepted

Note: The following works if $G$ is finite, but may fail in the infinite case; there are infinite groups in which there are $p$-Sylow subgroups $P$ and $P'$ for which no automorphism (inner or outer) of $G$ maps $P$ to $P'$.

Let $K$ be a $p$-Sylow subgroup of $H$ (we know it exists, though it may be trivial). Then $K$ is a $p$-subgroup of $G$ (even if $K={e}$) and by the Sylow Theorems, is contained in some Sylow $p$-subgroup $Q$ of $G$. By the Sylow Theorems, $Q$ and $P$ are conjugate. Now just verify that $Q\cap H = K$.

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Thanks, this works! I've only one question, and I'm sorry for all the trouble: Do all p-Sylow subgroups are conjugate even if G is infinite? (if it isn't to much trouble could you mention a book that proves this?) Thank you so much for your time. –  Chu Dec 29 '10 at 18:52
    
This apparently is not true in general. See mscand.dk/article.php?id=1527 for a counterexample. –  Hans Parshall Dec 29 '10 at 19:02
    
You're right, so the solution above doesn't work :(. Could it be possible the result is only valid for $G$ finite? –  Chu Dec 29 '10 at 19:08
    
@Chu: Apparently not, as Hans notes. Please un-accept and I'll see if there is a way around it in the infinite case. –  Arturo Magidin Dec 29 '10 at 19:10
    
@Chu: Are you positive the problem asks you to consider possibly infinite groups? –  Arturo Magidin Dec 29 '10 at 19:12
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Let $G$ be the direct product of countably many copies of the dihedral group $D$ of order 6 (or, if you prefer, $D$ is the symmetric group $S_3$).

We can construct a Sylow $2$-subgroup of $G$ by choosing Sylow $2$-subgroups of each of the direct factors of $G$, and taking their direct product. Since $D$ has three Sylow $2$-subgroups, $G$ has uncountably many Sylow $2$-subgroups, so they cannot all be conjugate in the countable group $G$.

If we let $P$ and $H$ be non-conjugate Sylow $2$-subgroups of $G$, then there is no $g \in G$ such that $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.

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Nice. Thanks for the example. –  Arturo Magidin Dec 30 '10 at 17:02
    
+1 - I like this better than the counterexample I found. Thank you. –  Hans Parshall Dec 30 '10 at 17:24
    
+1 ! this is a great example! thank you –  Chu Dec 30 '10 at 20:21
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Well, we have $\vert P \vert =p^n$ where $\vert G \vert = p^n m$ and $(p,m)=1$. What can be said about $\vert H \cap P \vert$ (or, for that matter, $\vert H \cap Q \vert$ for any Sylow p-subgroup $Q$ of $G$)?

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Jack, you meant that $(p,m)=1$ right? –  Asaf Karagila Dec 29 '10 at 17:36
    
Indeed I did. Thanks for catching that! It's been fixed. –  user5137 Dec 29 '10 at 17:39
    
thanks for your time :), one of the problems I have with this is that $G$ isn't necesarily finite. –  Chu Dec 29 '10 at 17:49
    
Ah, in that case, we know that for all $x\in P$, $\vert x \vert=p^n$ for some $n\geq 0$. What does that tell us about elements of $H\cap P$? –  user5137 Dec 29 '10 at 18:11
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Let $P'$ be any Sylow $p$-subgroup of $H$. Then there is a maximal $p$-subgroup of $G$ containing it. What is it?

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I'm a bit confused. Could it be possible that $H$ has no p-subgroups? –  Chu Dec 29 '10 at 18:21
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@Chu: The trivial subgroup is a p-subgroup. –  Hans Parshall Dec 29 '10 at 18:30
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