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Corollaries of Lagrange's Theorm

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Not a homework question actually. I'm having trouble understanding some of the corollaries to Lagrange's theorem.

    Theorem: Let H b e a subgroup of a finite group G. then |H| divides |G|

    Corollary 1: if g is an element of a finite group G, then |g| divides |G|.
    proof: the cyclic subgroup |H| = <g> generated by g has |H| = |g|.
    question: how do we know such a cyclic subgroup H exists as required by the proof?

    Corollary 2: If p is a prime, then every group G of order p is prime.
    proof: write H = <g>. Then |H| divides |G| so |H| is 1 or |H| = p = |G|.
    question: Again, how do we know that such a cyclic subgroup H exists in the first place?
     
  2. jcsd
  3. Aug 5, 2014 #2

    jbunniii

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    What is your definition of <g>?

    I assume you mean "is cyclic".
    Same question as above: what is your definition of <g>?
     
  4. Aug 5, 2014 #3
    <g> is defined a subgroup of G where g is an element of G and all the elements of <g> are {1, g, g^2, g^3, g^(n-1} where |G| = n. I believe.

    Yes I meant to say "is cyclic" not "is prime" my bad.
     
  5. Aug 5, 2014 #4

    jbunniii

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    So far so good.
    ##|G| = n## is only true if ##G## is cyclic (and finite), and ##g## is a generator for ##G##. More generally, ##\langle g\rangle## might be a proper cyclic subgroup of ##G##. However, the key is that it's a subgroup. So what does Lagrange's theorem tell you about the size of ##\langle g\rangle##?
     
  6. Aug 5, 2014 #5
    so this corollary only works if G is cyclic thus has a cyclic subgroup? or this corollary only works assuming there is a cyclic subgroup <g> of G? I feel like something is being assumed that the corollary statement didn't outright state.
     
  7. Aug 5, 2014 #6

    jbunniii

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    ##\langle g \rangle## is always a cyclic subgroup, even if ##G## is not.
     
  8. Aug 5, 2014 #7
    so every group has cyclic subgroups... I guess that's obvious now that I think about it. thanks a ton man, i'm new to this stuff X_X
     
  9. Aug 5, 2014 #8

    jbunniii

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    Yes, that's right. Every element ##g## generates a cyclic subgroup ##\langle g \rangle##, but not necessarily distinct subgroups - some elements might generate the same subgroup. (For example, ##\langle g \rangle = \langle g^{-1} \rangle##.)

    Since ##\langle g \rangle## is a subgroup, Lagrange's theorem constrains its size to be a divisor of ##|G|##.

    The converse is not true in general: if ##d## is a divisor of ##|G|##, there isn't necessarily a subgroup (cyclic or not) of order ##d##.

    However, pretty soon you will probably learn Cauchy's theorem, which says that if ##p## is any prime divisor of ##|G|##, then ##G## has a cyclic subgroup of order ##p##.
     
    Last edited: Aug 5, 2014
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