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Finite group with two prime factors

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to prove the following:

    Let [itex]G[/itex] be a finite group and let [itex]\{p,q\}[/itex] be the set of primes dividing the order of [itex]G[/itex]. Show that [itex]PQ=QP[/itex] for any [itex]P[/itex] Sylow [itex]p[/itex]-subgroup of [itex]G[/itex] and [itex]Q[/itex] Sylow [itex]q[/itex]-subgroup of [itex]G[/itex]. Deduce that [itex]G=PQ[/itex].

    2. Relevant equations
    The set [itex]PQ=\{xy: x \in P \text{ and } y \in Q\}[/itex]


    3. The attempt at a solution
    I know that [itex]PQ=QP[/itex] means that I must prove [itex]PQ[/itex] is a subgroup of [itex]G[/itex]. Let [itex]r[/itex], [itex]s[/itex] be elements of [itex]PQ[/itex]. Then [itex]r=x_{1}y_{1}[/itex] and [itex]s=x_{2}y_{2}[/itex].

    Now [itex]rs^{-1}=x_{1}y_{1}(x_{2}y_{2})^{-1}=x_{1}y_{1}y_{2}^{-1}x_{2}^{-1}[/itex] but I am not sure whether this is an element of [itex]PQ[/itex] or not.

    The second part by the following:
    [itex]|G|=|P||Q|=\frac{|P||Q|}{|P \cap Q|}=|PQ|[/itex] as [itex]P[/itex],[itex]Q[/itex] are Sylow for distinct primes of [itex]G[/itex] and [itex]P \cap Q=\{1\}[/itex]. This imply that [itex]G=PQ[/itex] as [itex]PQ[/itex] is a subgroup of [itex]G[/itex] with the same order of [itex]G[/itex].

    Thanks in advance.
     
  2. jcsd
  3. Nov 20, 2012 #2

    jbunniii

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    What happens if you reverse the roles of P and Q?
     
  4. Nov 20, 2012 #3
    In case [itex]PQ[/itex] is a subgroup of [itex]G[/itex], then [itex]PQ=QP[/itex].
     
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