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Group theory help!

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    Let G be a nontrivial group (remember that the trivial group is the group with only one element, so a
    nontrivial group is a group with at least 2 elements). Suppose that the only subgroups of G are G and
    {1}. Prove that G is cyclic and finite, and prove that the number of elements in G is a prime number.

    I can do the first part but not the second in bold, help would be appreciated.
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    If it isn't prime, can you show that it has some subgroups other than those stated?
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    Have a look at Sylow's theorems.
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    (Original post by Gimothy)
    Have a look at Sylow's theorems.

    wouldn't lagrange suffice?
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    Suppose G is infinite, and split it into two cases: if G is finitely-generated then it contains an element of infinite order; and if G is infinitely-generated then it contains infinitely many generators (duh). In each of these cases you can find a proper subgroup by choosing every second power or every second generator, say. I'll let you fill in the details.

    Now suppose G is finite. If it is generated by more than on element, can you find a proper subgroup? What about if it is generated by precisely one element which has composite order?

    (Original post by ben-smith)
    wouldn't lagrange suffice?
    No. That shows that if H \le G then \left| H \right| divides \left| G \right|; it doesn't show that if n divides \left| G \right| then there is a subgroup of order n. (And rightly so -- this isn't true!)
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    (Original post by nuodai)
    No. That shows that if H \le G then \left| H \right| divides \left| G \right|; it doesn't show that if n divides \left| G \right| then there is a subgroup of order n. (And rightly so -- this isn't true!)
    ooops, right you are
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    (Original post by Hopple)
    If it isn't prime, can you show that it has some subgroups other than those stated?
    I think I might be on the right tracks if I start by assuming if

    G=<g>, then write the identity element as g^(a) for some integer a and then do some more stuff. Will get back to you soon.

    Ok, I think that g^n=e must be the case. And because of this if n=ab a,b>1, we can take any (g^a), (g^2a),...,(g^ba) will form a subgroup. Think need to prove this a bit more rigorously but I'm starting to see it. If I'm on the wrong tracks please tell me.
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    Thanks TSR.
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    (Original post by ben-smith)
    wouldn't lagrange suffice?
    Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.
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    (Original post by Gimothy)
    Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.
    But it's pretty much overkill here, I think.
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    (Original post by Gimothy)
    Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.
    Does Sylow obviously cover the case where the order of G is p^n with p prime and n > 1 ? I'd use Cauchy's theorem instead. It's lighter and covers this case cleanly.
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    (Original post by SsEe)
    Does Sylow obviously cover the case where the order of G is p^n with p prime and n > 1 ? I'd use Cauchy's theorem instead. It's lighter and covers this case cleanly.
    Isn't Cauchy's theorem just a trivial consequence of Sylow's theorems?
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    Am I being stupid, or did Nuodai answer this completely without using Sylow/Cauchy/etc?
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    Every subgroup of a cyclic group is normal (because a cyclic group is abelian) and so appeal to the classification theorem for finite simple groups.
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    (Original post by DFranklin)
    Am I being stupid, or did Nuodai answer this completely without using Sylow/Cauchy/etc?
    Nuodai is workin in ZFC, pretty sure the OP wanted a solution working within ZF.
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    (a) I'm not sure why you think that, and (b) surely ZF v.s. ZFC only arises in the case G is infinite and since Cauchy/Sylow are for finite groups they can't help anyhow?
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    Srs bsnizz.

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