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Group theory, Lagrange theorem

  1. Oct 25, 2009 #1
    Dear all,
    The question Ive been struggling with is supposed to be solved using the way Lagrange's thm was proven( with number of cosets and stuff). However, it remains a mystery how to do it:

    Let G be a finite group and H<G with |G|=m|H|. Proof that

    [tex] g^{m!} \in H, \forall g \in G[/tex]
  2. jcsd
  3. Oct 25, 2009 #2
    Using Lagrange's theorem on your expression we get [itex][G : H] = |G|/|H| = m[/itex] so there are m cosets of H in G. Consider:
    Can they all be different? If not then we have an integer [itex]n \in \{1,2,\ldots,m\}[/itex] such that [itex]g^n \in H[/itex]. Now since H is a group and [itex]m!/n[/itex] is an integer, then H contains:
    [tex]\left(g^n\right)^{m!/n} = g^{m!}[/tex]
  4. Oct 25, 2009 #3
    Ok, thanks, however, it's still a bit blurry to me why the fact that they can't be all the same implies that there is such an integer n..
  5. Oct 25, 2009 #4
    Since they are not all the same we can find two that are equal. Let these be [itex]g^a H[/itex] and [itex]g^b H[/itex] with [itex]a,b \in \{0,1,\ldots,m\}[/itex]. Assume without loss of generality that a < b, then we have:
    [tex]g^b H = g^a H[/tex]
    which implies
    [tex]g^{b-a} H = g^{a-a}H = H[/tex]
    which is equivalent to [itex]g^{b-a} \in H[/itex].

    Since [itex]b \leq m[/itex] and a is non-negative [itex]b-a\leq m[/itex] and since b>a we have 0<b-a which together gives us [itex]0 < b-a \leq m[/itex] so n=b-a is the number we wanted to prove existed.
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