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I run into a problem when I try to solve a question, show $D_{2^{n+1}}/Z(D_{2^{n+1}})\cong D_{2^n}$. Note: $D_n$ is the dihedral group of order 2n. I have shown $D_{2^{n+1}}/Z(D_{2^{n+1}})\cong D_{2^{n+1}}/Z_2$. But I am not sure how to show $D_{2^{n+1}}/Z_2\cong D_{2^n}$

Thank you.

mathlad
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    “$D_{2^{n+1}}/Z_2$“ does not quite make sense, since $Z_2$, the integers modulo $2$, are not a literal subgroup of $D_{2^{n+1}}$. It is isomorphic to a particular subgroup (in fact, to many, but just to one normal one). You really need to know who $Z(D_{2^{n+1}})$ is as a specific subgroup, and use that. – Arturo Magidin Mar 31 '21 at 03:44
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    As to how to prove it, it depends on exactly how you “know” the dihedral groups. By a presentation? As specific subgroups of symmetric/permutation groups given by generators? As “group of symmetries” of specific objects? – Arturo Magidin Mar 31 '21 at 03:45
  • thank you for the comments, i used <$r^{2^n}$>, and it worked out. – mathlad Mar 31 '21 at 05:51
  • The isomorphism holds more generally for $D_m$, and $D_{2m}/Z$. In your case, $m=2^n$. – Dietrich Burde Mar 31 '21 at 08:23

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