For all $n > 0$ express $D_{2n}$ as a semidirect product $\mathbb Z_n \rtimes_\theta \mathbb Z_2$, finding $\theta$ explicitly.

I am not sure how to go about finding $\theta: \mathbb Z_2 \to \mathrm{Aut}(\mathbb Z_n)$ explicitly.

The map is determined by where $1 \in \mathbb Z_2$ maps to. Based on playing with $\mathbb Z_2 \to \mathrm{Aut}(\mathbb Z_4)$, I predict $\theta$ will map to the automorphism sending $1 \mapsto -1$ in $\mathbb Z_n$.

However, I don't want to just guess where to send $1$ to.

How would I begin to systematically find $\theta$ explicitly?


There is one very simple automorphism of $\Bbb Z_n$ (for $n>2$) which has order 2 and is described basically the same way no matter what $n$ is. That's what $\theta(1)$ is.

And guessing isn't that bad. The real work lies in showing that $\theta$ works anyways, not in finding it.

  • $\begingroup$ I know it has to be the inverse automorphism. But how would one go about making sure that this is where $1$ should map to? $\endgroup$ – Al Jebr Jul 7 at 23:11
  • $\begingroup$ @AlJebr You check that by the definition of semidirect product, that $\theta$ does indeed give the dihedral group. $\endgroup$ – Arthur Jul 7 at 23:24
  • $\begingroup$ Do you mean constructing an isomorphism of the semi-direct product $\mathbb Z_n \rtimes_\theta \mathbb Z_2$ with respect to that $\theta$ to $D_{2n}$? $\endgroup$ – Al Jebr Jul 7 at 23:31
  • $\begingroup$ @AlJebr Yes, that's what you're asked to do. $\endgroup$ – Arthur Jul 8 at 0:07
  • $\begingroup$ To emphasize the point, sometimes finding the correct mathematical object requires an entirely unsystematic leap of human imagination, $\endgroup$ – Lee Mosher Jul 8 at 1:32

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