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Groups of prime power order

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Any help with this question would be great:

    G is a group such that |G| = pk, p is prime and k is a positive integer. Show that G must have an element of order p.

    The hint is to consider a non-trivial subgroup of minimal order.

    2. Relevant equations


    3. The attempt at a solution

    Can I use Lagrange to say that there must exist a subgroup H of order pm with m<k? Or even a subgroup H of order p? Although I'm not sure how to justify this. Then if H was cyclic then there must be an element of order p? Again, this is just a guess.
  2. jcsd
  3. May 1, 2010 #2
    You are right to suspect that Lagrange's theorem cannot be used to justify that there exist subgroups of certain sizes. It can only be used directly to justify that there do not exist subgroups of certain sizes. Finding a (cyclic) subgroup H with p elements is enough, if you can do so, because the generator of H would have order p. (Why?) How can you find cyclic subgroups in G?
  4. May 1, 2010 #3
    I don't know?

    All I know is that G contains p^k elements. I'm a bit confused about how to deduce anything else about the group.
  5. May 1, 2010 #4
    Right, since we don't know anything else about this group, let's look at the elements of it. What order can these elements have?
  6. May 1, 2010 #5
    The order of the elements must divide p^k

    So they could be order p, and then p^t where t divides k.
  7. May 1, 2010 #6
    So if there is an element of order p, then we are done. Suppose not. Then, every (nonidentity) element has order p2, p3, ..., or pk. Now, remember that a cyclic group is generated by one element; a cyclic subgroup is generated by one element of this group. What do these cyclic subgroups look like?
  8. May 1, 2010 #7
    <p> = {p^k : k is an integer}

    This looks like the list you have just given.
  9. May 1, 2010 #8
    Care! You only need t<=k.
    1,3,9,27 all divide 27 for instance.
    That is 3t with t=0,1,2,3 all divide 3k with k=3.
  10. May 1, 2010 #9
    Yes, that's the definition of a cyclic subgroup for an element p. Let's use the letter g to denote an arbitrary element from now on to avoid confusion with the prime p. <g> is defined to be all the powers of g. But since g is an element of a finite group G, the cyclic subgroup generated by g must have finite order. In particular, the order is the order of g, which is the generating element.

    By assumption g has order pm for some 1 < m <= k, so <g> has order pm. Is there a natural nontrivial subgroup of <g> that has order less than pm?

    EDIT: I need to step our for a few hours. If you haven't figured it out, I'm sure someone else will pick up from where I left off. What I am trying to illustrate is the way I've come to reason about groups when I see such a problem. Thinking about orders of group elements and orders of groups. Thinking about what if there were no element of order p; what contradiction can arise? What particular subgroups are easy to identify, e.g. cyclic subgroups? Granted I am taking algebra this semester as well, and my viewpoint is not very advanced, but I hope that this helps a bit more than just giving you the answer.
    Last edited: May 1, 2010
  11. May 1, 2010 #10
    So is <g> = {e, g, g^2, ..., g^(p^m -1)}?

    I'm still a bit stumped as to how to find a subgroup.
  12. May 1, 2010 #11
    Yes, that lists explicitly the elements of <g>. We know <g> is a cyclic subgroup of G with pm elements. What possibly subgroups of this cyclic group are there? Such a nontrivial subgroup must include at least one nonidentity element. It must then contain all powers of that element by closure. Is there a nice choice for this first element that leads to a smaller subgroup? What sizes can you make these subgroups?
  13. May 1, 2010 #12
    What if we chose <g^p> ?

    I'm still not too sure, but ready to quite for the night now too. Thank you for all your help :)
  14. May 1, 2010 #13
    You are almost there. What is the order of <gp>? What does that tell you about the order of the element gp? Is there another element that is better to use?
  15. May 1, 2010 #14
    Is the order of <g^p> m? Then the order of g^p is also m?
  16. May 1, 2010 #15
    Careful there. Is (gp)m equal to gpm?
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