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How did the author get that $L=(L \cap H)(L\cap K)$ in Lemma $5$ below.

Remark: All the groups here are finite. $H$ permutes (commutes) with $K$ means $HK=KH$ where $H$ and $K$ are subgroups of some finite group $G$.

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Thanks in advance.

  • 1
    What primes could possibly divide $|L:(L \cap H)(L \cap K)|$? None that lie in $\pi(H)$ and none that do not lie in $\pi(H)$.2012-11-19
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    @DerekHolt: There is no other than one. But, I do not know how to get that from analyzing $|L:(L \cap H)(L \cap K)|$.2012-11-20
  • 0
    You know that $|L:L \cap K|$ is divisible only by primes in $\pi(H)$ and you know also that $|L:L \cap H|$ is divisible only be primes that do NOT lie in $\pi(H)$.2012-11-20
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    @DerekHolt: I understand your last comment. Sorry, I still don't get it.2012-11-20
  • 0
    If $|L:(L \cap H)(L \cap K)|$ is not divisible by any primes, then it must equal 1.2012-11-20
  • 0
    @DerekHolt: $(L \cap H)(L \cap K)|=\frac{|L||L \cap H \cap K|}{|L \cap H||L \cap K|}$ How is this not divisible by any prime?2012-11-20
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    @DerekHolt: $|L:(L \cap H)(L \cap K)|=\frac{|L||L \cap H \cap K|}{|(L \cap H)||(L \cap K)|}$, How is this not divisible by any prime? I made a mistake in the previous comment.2012-11-21
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    I have said twice already that it is not divisible by any primes that are in $\pi(H)$ and it also not divisble by any primes that are NOT in $\pi(H)$!2012-11-21
  • 0
    @DerekHolt: Thank you very much.2012-11-21

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