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

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.
• 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? – Al Jebr Jul 7 at 23:11
• @AlJebr You check that by the definition of semidirect product, that $\theta$ does indeed give the dihedral group. – Arthur Jul 7 at 23:24
• 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 $D_{2n}$? – Al Jebr Jul 7 at 23:31