Constructing the semidirect product $C_{2^n}\rtimes C_2$. Starting from the group of automorphisms of $C_{2^{n}}$ find the automorphism $x$ of order 2 (with $n\geq 3$) and building the semidirect product $C_2$ and $C_{2^{n}}$ by $y$, where $y$ is an application of $C_2$ into $x$.
Can you help me?
 A: There are three distinct automorphisms of order $2$ of $C_{2^{n}}$, for $n \ge 3$, and these are
$$
x \mapsto x^{-1},
\qquad
x \mapsto x^{-1 + 2^{n-1}},
\qquad
x \mapsto x^{1 + 2^{n-1}}.
$$
The corresponding semidirect products are called respectively, the dihedral group, the quasi-dihedral (or semi-dihedral) group, while the third one (poor guy) apparently has no established name.
This depends on a peculiarity of the prime $2$ when it comes to the unit group of the cyclic group $C_{p^{n}}$, with $p$ a prime. For $p$ odd the unit group is isomorphism to $C_{p-1} \times C_{p^{n-1}}$. For $p = 2$, and $n \ge 3$, it is isomorphic to $C_{2} \times C_{2^{n-2}}$, hence the three automorphisms of order $2$.
A: $\operatorname{Aut}(C_{2^n})$ is isomorphic to the unit group $C_{2^n}^\star$.  Finding an automorphism of order $2$ in there isn't hard: you just need something such that $n^2\equiv 1 \pmod{2^n}$.  That's easy, though, it's just $x\mapsto -x$.  So you will form $C_{2^n}\rtimes C_2=\langle x \rangle \rtimes \langle y \rangle$ by letting $y$ act on $x$ by inversion.  A group presentation would be $\langle x,y | x^{2^n},y^2,y^{-1}xy=x^{-1} \rangle$.  But this is just the dihedral group $D_{2^{n+1}}$!
