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Group Theory - Prime Order + Subgroups

  1. Mar 23, 2006 #1
    Problem:

    "Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime. (Do not assume at the outset that G is finite)."

    Basically, I'm pretty sure I can do this problem. I'm just unsure of how to prove that all infinite groups have some proper nontrivial subgroups.

    I also thought about just proving that all infinite groups and all groups with order not prime implies they have proper nontrivial subgroups.

    Either way I'm stuck on setting this up, any help would be of great help.
     
  2. jcsd
  3. Mar 23, 2006 #2

    Hurkyl

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    One step at a time.

    The first thing I did when I read your question is that I started imagining infinite groups. Z was the first one that came to mind. It's easy to find a proper nontrivial subgroup of Z, right? So my next thought was if the same idea could apply to any infinite group!


    Incidentally, this is a good "formula" -- it's a useful thing to do on just about any problem: try to study special cases to see if they give you insignt.
     
    Last edited: Mar 23, 2006
  4. Mar 23, 2006 #3

    matt grime

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    I would think as follows:

    let g be a nonidentity element (such exists), if g has finite order and g is not all of the group we have a subgroup, if g is finite order and is all of the group then..... and if g is of infinite order I can see there is a proper subgroup generated by...

    just to make sure you've got all possible cases covered.
     
  5. Mar 23, 2006 #4
    I'm confused how either approach proves that G must be prime. I'm just not seeing the trick obviously.

    Given a infinite group I can understand that it would be easy to find a subgroup, such as the cyclic subgroup 2 in Z. But, I'm unsure how I could transfer this into a general proof about all infinite subgroups.

    As for the second idea, it seems like a good start. It incorporates part of the question information which is always a good thing in my experience :P.

    If g has finite order and g is not all of the group we have a subgroup. - I'm not sure how this ties in, G could be finite or infinite, no way to tell.

    If g is finite order and is all of the group then... its a cycllic group? Still unsure how this specifies what kind of group G we have.
     
  6. Mar 23, 2006 #5

    Hurkyl

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    We both have (more or less) the same proof -- we just started from opposite ends of the problem!


    I'm confused by your second sentence. You just said that you think it's easy to find a proper nontrivial subgroup of an infinite group. Isn't that exactly what you wanted to show?

    Does it matter?


    I'm confused by this one too. You say that G is a cyclic group, and then say that you're unsure how that tells you what kind of group G is.


    P.S. when matt said "g is not all of the group", he didn't mean g, he meant the subgroup generated by g
     
    Last edited: Mar 23, 2006
  7. Mar 24, 2006 #6
    Matt: After re-reading your suggestion I can see how it ties in. Thanks. One question: How do we know if g has infinite order that its a proper subgroup of G? If it has infinite order and contains all of G then wouldnt it be considered a trivial subgroup of G since its not a subgroup at all but a generator of G?

    Well... I have done every case were g has finite order. I'm still unsure how to proceed for the case where it has infinite order. I'm unsure how we know it generates a proper subgroup of G.
     
    Last edited: Mar 24, 2006
  8. Mar 24, 2006 #7
    I think I have my missing piece, tell me if its wrong.


    If g has infinite order and generates all G. Then the element h = g^2 generates a proper subgroup of G. (All even powers of g).

    If g has infinite order and does not generate all G then <g> is a subgroup of G.

    If g has finite order and generates all G. The order of G must be finite and therefore the order of every element must divide the order of the group. Since the only elements of order 1 is the identity and this group has more then one element the order must be prime or else there will be an element h in G that generates a cyclic subgroup. Note: if no h exists and g and the identity are the only elements then the order of G is 2 which is prime.

    If g has finite order and does not generate all G we have a subgroup.

    Nvm: I just remembered the fundamental law of cyclic subgroups ^_^. For any divisor of the order of G there is exactly one subgroup of that order.
     
    Last edited: Mar 24, 2006
  9. Mar 24, 2006 #8

    matt grime

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    This bit is very unnecessarily opaque:

    "If g has finite order and generates all G. The order of G must be finite and therefore the order of every element must divide the order of the group. Since the only elements of order 1 is the identity and this group has more then one element the order must be prime or else there will be an element h in G that generates a cyclic subgroup. Note: if no h exists and g and the identity are the only elements then the order of G is 2 which is prime."

    Perhaps your "nvm" was correcting that?

    Apologies for not using <g> is all of the group, it probably caused unnecessary confusion.
     
  10. Mar 25, 2006 #9
    Thanks for the tip either way. Gave me a great starting point for the problem.
     
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