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Math Help - Cyclic group

  1. #1
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    Cyclic group

    Dear all,

    Can I have a question

    Let G be a finite group with no subgroups apart from [IG] and G. Show that G is cyclic.

    I heard if A and B are both finite group and A is a subgroup of B, they are the same size.
    So in the question, can I assume G as B, and A as what?

    I am sorry to bother you.
    Thanks a lot.
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  2. #2
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    Re: Cyclic group

    Quote Originally Posted by yanirose View Post
    Let G be a finite group with no subgroups apart from [IG] and G. Show that G is cyclic.
    The facts are: the group is finite and has no non-trivial subgroups.
    From those two, prove that any non-identity in the group generates the whole group.
    Thanks from yanirose
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  3. #3
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    Re: Cyclic group

    Thanks a lot.

    Then how can I show that the number of elements in G is either 1 or a prime number?
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  4. #4
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    Re: Cyclic group

    Quote Originally Posted by Plato View Post
    The facts are: the group is finite and has no non-trivial subgroups.
    From those two, prove that any non-identity in the group generates the whole group.
    The order of any sub-group of a finite group divides the order of the group.
    Thanks from yanirose
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  5. #5
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    Re: Cyclic group

    Hi,
    I don't quite see how response 4 (Lagrange's Theorem) proves that the order of G is prime. So repeating:
    Let G be a finite group of order greater than one such that the only subgroups of G are G and <1>. Then G is cyclic of prime order.
    Proof.
    Recall: in any group G and $x\in G$, the order of the element $x$ is the order of the subgroup $<x>$. Since the order of G is greater than 1, G has non-identity elements. For any $x\in G$ with $x\neq1$, $<x>\neq1$ and so $<x>=G$. So $G=<x>$ is cyclic of order say $n>1$. Let $p$ be any prime divisor of $n$. Then either $<x^p>=G$ or $<x^p>=<1>$. But the order of the element $x^p$ is $n/p<n$ and so the order of $<x^p>$ must be 1. That is $x^p=1$ and so the order of x = p = the order of G.
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