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Math Help - D2n Isomorphic To Dn X Z2

  1. #1
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    D2n Isomorphic To Dn X Z2

    Show that D_{2n} \cong D_n \times \mathbb{Z}_2

    I'm using x, y to represent the elements in D_{2n} and a, b for those elements in D_n. My first idea was to do this:

    Map x^{2i} \in D_{2n} to (a^i, 0) \in D_n \times \mathbb{Z}_2 since that forms a cyclic subgroup of the correct order.

    Now I can't figure out where to go from here, or if my initial idea is even remotely in the ballpark.
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    Show that D_{2n} \cong D_n \times \mathbb{Z}_2

    I'm using x, y to represent the elements in D_{2n} and a, b for those elements in D_n. My first idea was to do this:

    Map x^{2i} \in D_{2n} to (a^i, 0) \in D_n \times \mathbb{Z}_2 since that forms a cyclic subgroup of the correct order.

    Now I can't figure out where to go from here, or if my initial idea is even remotely in the ballpark.
    This result is only true when n is odd. In that case, it is correct to map x^{2i} \in D_{2n} to (a^i, 0) \in D_n \times \mathbb{Z}_2. The correspondence is completed by mapping x^n \in D_{2n} to (e,1) \in D_n \times \mathbb{Z}_2, and y \in D_{2n} to (b,0) \in D_n \times \mathbb{Z}_2 (where  e is the identity element of D_n).
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