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Cyclic group has 3 subgroups, what is the order of G

  1. Feb 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose a cyclic group, G, has only three distinct subgroups: e, G itself,
    and a subgroup of order 5. What is |G|? What if you replace 5 by p where
    p is prime?

    2. Relevant equations


    3. The attempt at a solution
    So, G has three distinct subgroups. By Lagrange's theorem, the order of the subgroup has to divide the order of the group. So the order of G is a multiple of 5. If we let |G| = 5, then there are only two subgroups, G and e. So we try |G| = 10. Why couldn't 10 be correct? Am I neglecting the fact that G is cyclic? (I know that the answer is actually 25, but am not sure why).
     
  2. jcsd
  3. Feb 15, 2017 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    The wording of the problem implies that G can't have a subgroup of order 2.
     
  4. Feb 16, 2017 #3
    But if G has order 10 does that mean that it must necessarily have a subgroup of order 2 and a subgroup of order 5? I thought that lagranges theorem was just an if statement and not an if and only if statement.
     
  5. Feb 16, 2017 #4
    For cyclic subgroups, Lagranges Theorem does work both ways. If k divides |G|, then {e, a^k, a^(2k), .... a^(|G| -k }, is a subgroup of G
     
  6. Feb 16, 2017 #5

    fresh_42

    Staff: Mentor

    A cyclic group ##G## is Abelian, so all subgroups ##U## are normal subgroups and thus a direct factor. This means ##G/U## is also (isomorphic to) a subgroup ##V## of ##G##.
     
  7. Feb 16, 2017 #6

    pasmith

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    Homework Helper

    A finite group of even order must have an element of order 2: Partition the group into subsets [itex]\{a, a^{-1}\}[/itex]. Such a set contains two elements if [itex]a^2 \neq e[/itex] or one element if [itex]a^2 = e[/itex].

    One of these subsets is [itex]\{e\}[/itex] but the group order is even, so there must be at least one other subset consisting of a single element.
     
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