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Math Help - [SOLVED] Every Subgroup of Dn of Odd order is cyclic

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    Member Haven's Avatar
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    [SOLVED] Every Subgroup of Dn of Odd order is cyclic

    Prove that any subgroup of  D_n of odd order is cyclic.

    So far I have:
    Let H \leq D_n
    Let r denote a simple rotation such that r^n = e, and let f denote a vertical flip
    if |H| = 1, we're done since H = {e}, which is cyclic
    otherwise let |H| = 2k+1 for some non-negative k \in \mathbb{N}

    Consider a \in H.
    If a = r^m*f  for some m \in \mathbb{Z}, then ord(a) = 2. Since the order of any element must divide the order of the group, we have that a cannot be in H.

    So the only elements of H are of the form r^m.

    But I can't seem to show that H must be generated by an element of the form r^m. Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Haven View Post
    Prove that any subgroup of  D_n of odd order is cyclic.

    So far I have:
    Let H \leq D_n
    Let r denote a simple rotation such that r^n = e, and let f denote a vertical flip
    if |H| = 1, we're done since H = {e}, which is cyclic
    otherwise let |H| = 2k+1 for some non-negative k \in \mathbb{N}

    Consider a \in H.
    If a = r^m*f for some m \in \mathbb{Z}, then ord(a) = 2. Since the order of any element must divide the order of the group, we have that a cannot be in H.

    So the only elements of H are of the form r^m.

    But I can't seem to show that H must be generated by an element of the form r^m. Any help would be greatly appreciated.
    H \subseteq \langle r \rangle and we know that every subgroup of a cyclic group is cyclic.
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