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Math Help - Finite group with identity subgroup.

  1. #1
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    Finite group with identity subgroup.

    If you have a finite group, say G with no subgroups apart from {1G } and G. How would you go about showing that G is cyclic and that the number of elements in G is either 1 or prime?

    I know that the order of this group is 1,but how would you show that this makes it cyclic, by Lagrange's theorem?
    Last edited by chipette; February 18th 2010 at 08:33 AM.
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  2. #2
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    Do you agree that, by definition, a group G is cyclic iff there is some a in G such that <a>=G.

    I know that the order of this group is 1
    Since the order of a finite group is its cardinality, the one and only group of order 1 is the trivial group G=\{1_G\}. But there are non trivial groups with no subgroup other than \{1\} and themselves (i.e. with no non-trivial subgroup).


    So, consider a finite group G\neq\{1_G\} with no non-trivial subgroup, let's say G has order n\geq 2. You may know that if a prime p divides n, then G has an element of order p.
    Assume n is not a prime, and try to obtain a contradiction by proving G has a non-trivial subgroup.

    So far you'll have that a finite group with no non-trivial subgroup has order 1 or a prime. If it is 1, do you agree that G=\{1_G\}=<1_G> therefore G is cyclic, because its only element is a generator. If it is not 1, take an element of order strictly greater than 1 (justify its existence) and consider the subgroup it spans. Because of the hypothesis, since it is different from \{1_G\}, it must be the whole group, hence G is cyclic.

    Note that we did not to use that order (when different from 1) is prime; that being said a similar argument shows that any group with order a prime has no non-trivial subgroup (and is cyclic).

    Conclusion: A group has no non-trivial subgroup iff its order is 1 or a prime.
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