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A detail in a proof about isomorphism classes of groups of order 21

  1. Mar 20, 2014 #1
    1. The problem statement, all variables and given/known data

    While reading through my textbook on abstract algebra while studying for a test, I ran across the following statement:

    There are two isomorphism classes of groups of order 21: the class of ##C_{21}##, and the class of a group ##G## generated by two elements ##x## and ##y## that satisfy the relations ##x^7=1##, ##y^3=1##, ##yx=x^2y##

    along with a proof of this statement. The proof looks fairly reasonable, only there's a bit I don't quite understand:

    "The Third Sylow Theorem shows us that the Sylow 7-subgroup ##K## [existence and uniqueness guaranteed by the Third Sylow Theorem] must be normal ..."

    Specifically, why must ##K\trianglelefteq G##?

    2. Relevant equations

    Third Sylow Theorem: If ##G## is a group of order ##n<\aleph_0##, where ##n=p^e\cdot m## for some ##e>0## and ##\lnot\left(p|m\right)##, for some prime ##p##, and ##s## is the number of Sylow ##p##-subgroups of ##G##, then ##s|m## and ##s\equiv1\pmod p##.

    3. The attempt at a solution

    Fairly nonexistent; I can't think of any reason this subgroup must be normal.

    (If this post would be more appropriate in the linear & abstract algebra section, feel free to move it there.)
     
  2. jcsd
  3. Mar 20, 2014 #2

    Dick

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    Use what they are saying about ##s## to show there must be only one 7-subgroup. So?
     
  4. Mar 20, 2014 #3
    There must be a 7-subgroup; I just don't see why it must be a normal subgroup.
     
  5. Mar 20, 2014 #4

    Dick

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    Use the definition of normal. There is only ONE 7-subgroup.
     
  6. Mar 20, 2014 #5
    I'm used to normal being defined as "closed under conjugation by elements of G." Is there a different or equivalent definition I'm missing or something else obvious which makes clear that only one 7-subgroup existing means said subgroup is normal? (Knowing me, most likely.)
     
  7. Mar 20, 2014 #6

    jbunniii

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    If you conjugate a subgroup, the result is another subgroup of the same size. So if there's only one subgroup of order 7...
     
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