sylow subgroup of a subgroup

1,775

Solution 1

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 $Q\cap H = K$.

Solution 2

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 $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.

Solution 3

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 $G$)?

Solution 4

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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Updated on October 15, 2022

Comments

  • Sak
    Sak about 1 year

    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.

    • Sak
      Sak almost 13 years
      $G$ need not be finite.
    • Mariano Suárez-Álvarez
      Mariano Suárez-Álvarez almost 13 years
      edit the question and add the information that $G$ need not be finite there.
  • Asaf Karagila
    Asaf Karagila almost 13 years
    Jack, you meant that $(p,m)=1$ right?
  • Admin
    Admin almost 13 years
    Indeed I did. Thanks for catching that! It's been fixed.
  • Sak
    Sak almost 13 years
    thanks for your time :), one of the problems I have with this is that $G$ isn't necesarily finite.
  • Admin
    Admin almost 13 years
    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 $H\cap P$?
  • Sak
    Sak almost 13 years
    I'm a bit confused. Could it be possible that $H$ has no p-subgroups?
  • Hans Parshall
    Hans Parshall almost 13 years
    @Chu: The trivial subgroup is a p-subgroup.
  • Sak
    Sak almost 13 years
    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.
  • Hans Parshall
    Hans Parshall almost 13 years
    This apparently is not true in general. See mscand.dk/article.php?id=1527 for a counterexample.
  • Sak
    Sak almost 13 years
    You're right, so the solution above doesn't work :(. Could it be possible the result is only valid for $G$ finite?
  • Arturo Magidin
    Arturo Magidin almost 13 years
    @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
    Arturo Magidin almost 13 years
    @Chu: Are you positive the problem asks you to consider possibly infinite groups?
  • Sak
    Sak almost 13 years
    This was on an exam I took recently and I couldn't solve this one. It's phrased without any assumptions on $G$'s finiteness so I assumed it could be infinite.
  • Arturo Magidin
    Arturo Magidin almost 13 years
    @Chu: Let me ask you this: How did you define "$p$-Sylow subgroup"? Was it defined as a maximal $p$-subgroup, or was it defined as a $p$-subgroup $P$ such that the order and the index of the subgroup were relatively prime, or as a subgroup whose order is the maximum power of $p$ that divides the order of $G$? The definition, not any equivalences proven later.
  • Sak
    Sak almost 13 years
    We defined it as a p-subgroup P such that the order and the index are coprime. So now I see this was intended for the finite case right? I'm sorry for all the trouble. I'll edit the question and re-accept your answer. Thanks again :).
  • Arturo Magidin
    Arturo Magidin almost 13 years
    @Chu: Yes, if you defined it as a subgroup in which the order and index are coprime, then your definition, a fortiori assumes that the group $G$ is finite (otherwise, the definition does not make sense since at least one of the order and index will necessarily be infinite). One can instead define Sylow subgroups as "maximal $p$-subgroups", but then they don't always exist (in the infinite case), and as B.H. Neumann's paper shows, we can have weird things happening in the infinite case.
  • Arturo Magidin
    Arturo Magidin almost 13 years
    Nice. Thanks for the example.
  • Hans Parshall
    Hans Parshall almost 13 years
    +1 - I like this better than the counterexample I found. Thank you.