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There is a semi direct product if you go $\theta: Q \rightarrow Aut(H)$.

$H = C_{17}, Q = C_2$.

$Aut(H)\cong C_{16}$.

From here, how do I construct the semi direct products?

I said, in $C_{16}$, there are two elements of order that divide $2$, which are $1$ and $2$. Therefore there are two semi direct products. Also $17 \times 2 = 34$ tells me that in both semi direct products there will be $34$ elements.

How do I construct the semi direct products from here though?

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  • $\begingroup$ You know that there at most two groups of order $34$, and you know that the dihedral group of order $34$ and $C_{34}$ are groups of order $34$. Hence, it must be those two! $\endgroup$ – Daniel Montealegre Dec 12 '12 at 21:12
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There doesn't seem to be many options, does it? Putting $\,C_2=\langle a\rangle\,\,,\,\operatorname{Aut}(H)=\langle\phi\rangle\,$ , we get for the non-trivial semidirect product:

$$F:=C_2\to\operatorname{Aut}(C_{17})\cong C_{16}\Longrightarrow f(a):=\phi^8$$

where, of course (why "of course"?), $\,\phi^8\,$ is just inversion: $\,\phi^8(x):=x^{-1}\,\,\,\,\,\forall\,\,x\in C_{17}\,$ .

Thus, denoting $\,\phi^8(x)=f(a)(x)=:x^a\,$, you have the direct product

$$C_{17}\rtimes C_2\,,\,\,\,(x,y)(x',y'):=(x(x')^y\,,\,yy')$$

As for the trivial homomorphism $\,C_2\to\operatorname{Aut}(C_{17})\,\,,\,\,a\to \operatorname{Id.}=\text{the identity automorphism}$ , this gives us a direct product which is the only abelian group of order $\,34\,$ , the cyclic one, .

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