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    Prove cyclic

    Prove that every subgroup of D_n of odd order is cyclic
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    Quote Originally Posted by mandy123 View Post
    Prove that every subgroup of D_n of odd order is cyclic
    Remember that D_n = \{ a^ib^j| 0\leq i\leq n-1, 0\leq j\leq 1, ba=a^{n-1}b\}
    Notice that (a^kb)^2 = a^k ba^k b = a^k a^{-k} bb = b^2 = 1
    This means that each b,ab,...,a^{n-1}b have order 2.
    Thus, this subgroup cannot contain these elements for then Lagrange's theorem would imply 2 two divides its order.

    This means if H is a subgroup of D_n of odd order it contains the elements e,a,...,a^{n-1}. Thus, it is a subgroup of \left< a\right>. But every subgroup of a cyclic group is cyclic. Thus, H is cyclic itself.
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