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Homework Help: Induction problem

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose the G is a group of order p^n, where p is prime, and G has exactly one subgroup for each divisor of p^n. Show that G is cyclic.


    2. Relevant equations



    3. The attempt at a solution
    First, the inductive basis holds for n=1. Now for each k>=1, we assume that the claim is true for a group of order p^k where k < n. We shall show that the claim is also true for a group of order p^n.

    Note that in a general case, the group of order p^n is not necessarily abelian. However, by the given inductive hypotheis, we show that the above group G of order p^n is abelian. We know that Z(G) is non-trivial for every nontrivial finite p-group and is a subgroup of G.

    From this I am suppose to use the G/Z(G) theroem to show that G/Z(G) is cyclic and thus that G is abelian. However, I don't see how I this is true, for example what if |G|=p^3 and |Z(G)|=p, then G/Z(G)= p^2, which does not mean that G/Z(G) is cyclic. (If I remember right G/Z(G) has to be trivial cyclic)
    From this I can use the Fundamental Theorem of Finite Abelian Groups to get the G is isomorphic to Zp+Z(p^n-1) or Zp and then conclude that it is isomorphic to Zp.
     
  2. jcsd
  3. May 16, 2010 #2
    What are the divisors of [itex]p^{n}[/itex]?
     
  4. May 16, 2010 #3
    1, p, p to m where m divides n and p^n. We know that we can't have 1 because Z(G) needs more than one lement
     
  5. May 16, 2010 #4
    Every group has 2 subgroups, namely [itex]\{e\}[/itex] and [itex]G[/itex]. So, there is always a subgroup with 1 element.
     
  6. May 16, 2010 #5
    {e} is always a subgroup, but in are case Z(G) cannot equal e since are group G is of order p^n, where p is a prime.
     
  7. May 16, 2010 #6
    I'm sorry. Maybe it's trivial, but I am unfamiliar with the notation [itex]Z(G)[/itex]. What do you mean by it?
     
  8. May 16, 2010 #7
    Z(G) is the center of the group G, for a to be an element of Z(G) a has to commute with all elements of G. Z(G) can be trivial (i.e just e) in some cases, but in this one it can't since |G|=p^n where p is a prime (p0groups have non-trivial centers).
     
  9. May 17, 2010 #8

    Office_Shredder

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    You're probably supposed to use the inductive hypothesis on the quotient G/Z(G)
     
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