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G is cyclic and |G| = p^n, p is prime <=> H,K Subgroups, H⊆K or K⊆H

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the following conditions are equivalent for a finite group G:

    1.[itex]G[/itex] is cyclic and [itex]|G| = p^n[/itex] where [itex]p[/itex] is prime and [itex]n\geq 0[/itex]
    2.If [itex]H[/itex] and [itex]K[/itex] are subgroups of [itex]G[/itex], either [itex]H⊆K[/itex] or [itex]K⊆H[/itex].


    3. The attempt at a solution

    1 => 2.

    Let [itex]H,K[/itex] be subgroups of [itex]G = <g>[/itex] where [itex]o(g) = p^n[/itex]. We have [itex]H = <g^a>[/itex] and [itex]K = <g^b>[/itex] where [itex]a[/itex] and [itex]b[/itex] divide [itex]p^n[/itex]. Since [itex]p[/itex] is prime, [itex]a = p^s[/itex] and [itex]b = p^t[/itex]. If [itex]s \leq t[/itex], this means [itex]a|b[/itex] whence [itex]H⊆K[/itex]. Similarly, if [itex]b \leq a[/itex] we have [itex]K⊆H[/itex].

    Now I'm stuck at 2 => 1. Any help is appreciated :rolleyes:
     
  2. jcsd
  3. Jun 18, 2012 #2
    Maybe if you can use a fact like, for any integer that divides the group G, there is a subgroup of that order? Do you have a fact like that that you are permitted to use? Maybe only for prime divisors. That result I think comes sooner.

    In other words, a strategy that might have lead you to this is to think about the negation of 1, so what if the order of G is not a power of a prime. And how this might lead to subgroups which are not "one a subset of another".
     
  4. Jun 18, 2012 #3
    If [itex]H[/itex] and [itex]K[/itex] are subgroups of [itex]G[/itex], suppose [itex]H⊆K[/itex]. So we have that [itex]|H|[/itex] divides [itex]|K|[/itex], and they both divide [itex]|G|[/itex]. If it would happen that [itex]|G| = p^ka^r[/itex] where [itex]p, a[/itex] are different primes, then [itex]G[/itex] would have two subgroups [itex]M[/itex] and [itex]N[/itex] such that [itex]|M|[/itex] divides [itex]p[/itex], [itex]|N|[/itex] divides [itex]a[/itex], [itex]M ⊄ N[/itex] and [itex]N ⊄ M[/itex], thus contradicting the condition.
    So [itex]|G| = p^k[/itex] for a prime [itex]p[/itex] and an integer [itex]k[/itex].

    Is it correct?
    Now how do I show that [itex]G[/itex] is cyclic?
     
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