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Intersection of two subgroups trivial, union is the whole group

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##G## be a group of order ##n## where ##n## is an odd squarefree prime (that is, ##n=p_1p_2\cdots p_r## where ##p_i## is an odd prime that appears only once, each ##p_i## distinct). Let ##N## be normal in ##G##. If I have that ##|G/N|=p_j## for some prime in the prime factorization of ##n##, and ##|N|=\frac{n}{p_j}##, then are the following true?
    ##N \cap G/N = \{e\}##
    and
    ##N \sqcup G/N = G##



    2. Relevant equations



    3. The attempt at a solution
    For the first claim, if you take ##N \cap G/N## and obtain a non-trivial element, that element will generate ##G/N##. And since that element is also in ##N##, I believe that would mean ##G/N < N##, but by LaGrange's theorem, since their orders are relatively prime, ##G/N## cannot be a subgroup of ##N##. Thus, the intersection is trivial. I'm a bit stuck on getting started on the second condition.
     
  2. jcsd
  3. Dec 5, 2013 #2

    jbunniii

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    What does ##N \cap G/N## mean? ##N## is a subset of ##G##, but ##G/N## is not: it is a collection of subsets of ##G##. What do you mean by the intersection of these objects?
     
    Last edited: Dec 5, 2013
  4. Dec 5, 2013 #3
    Yes I apologize. When writing up the question, I got excited that this lemma would help me prove a bigger problem I had. Indeed, ##G/N## is a group since ##N\vartriangleright G##, but of course ##G/N## itself is not a subgroup of ##G##, which renders the lemma useless in my case, though, I do like the lemma. Thank you.
     
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