# 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?

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"... and building the semidirect product..." what? You have a dangling clause there. What is one supposed to do by building the semidirect product? –  Arturo Magidin May 26 '11 at 15:42
"Can you help me?" Not right now, because you didn't say what you need help with. If by "help" you mean solve the exercise for you, then again the answer is "no", because that's not the point of the exercise. By the way, "find the automorphism" should probably read "find an automorphism", since there are 3 of them. –  Alex B. May 26 '11 at 16:11
To get automorphisms of order 2, ref. "A classical Introduction to Modern Number Theory", chapter 4, and to construct semidirect products, up to isomorphism, refer "Groups and Representations"-Alperin,Bell; a section on "Semidirect Products". –  user8186 May 28 '11 at 5:15

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 $2$.
$\operatorname{Aut}(C_{2^n})$ 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 $D_{2^{n+1}}$!