1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.