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Abelian groups and exponent of a group

  1. Aug 16, 2012 #1
    Let [itex]p[/itex] be a prime. Let [itex]H_{i}[/itex], [itex]i=1,...,n[/itex] be normal subgroups of a finite group [itex]G[/itex]. I want to prove the following:
    If [itex]G/H_{i}[/itex], [itex]i=1,...,n[/itex] are abelian groups of exponent dividing [itex]p-1[/itex], then [itex]G/N[/itex] is abelian group of exponent dividing [itex]p-1[/itex] where [itex]N=\bigcap H_{i} ,i=1,...,n[/itex].

    Since [itex]G/H_{i}[/itex], [itex]i=1,...,n[/itex] are abelian groups, then [itex]G^{'}[/itex] (the derived subgroup of [itex]G[/itex]) is contained in every [itex]H_{i}[/itex], [itex]i=1,...,n[/itex]. Hence [itex]G^{'}[/itex] is contained in [itex]N[/itex]. Therefore [itex]G/N[/itex] is abelian. I do not know how to deal with the exponent.
    Thanks in advance.
  2. jcsd
  3. Aug 16, 2012 #2

    I think you only need the following two things:

    1) If A is an abelian group of exponent m, then the exponent of any subgroup of A and of any homomorphic image of A divides m

    [tex](2)\,\,\forall i\,\,,\,\, N \leq H_i\, \Longrightarrow \,\,\forall\,\, g\in G\,\,,\, m(g + N) = N\Longrightarrow mg \in N \leq H_i\Longrightarrow \,m(g + H_i) = H_i \Longrightarrow [/tex]
    [tex]\Longrightarrow\, m\,\, \text{divides the exponent of}\,\, G/H_i\,\, .[/tex]

  4. Aug 16, 2012 #3
    thank you very much.
  5. Aug 16, 2012 #4
    Excuse me, how did you know that m is less than the exponent of G/Hi?
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