1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook