Two questions on calculating semi direct products 1) First of all, I have to construct the semi direct product "from H to Q", it is written has $H \lambda Q$ in my notes. I have $H = C_{17}, Q = C_2$.
So I did $Aut(C_{17}) = C_{16}$ and then you can see that 2 divides 16 and so you want to work out how many elements of order 2 there are in $C_{16}$. Doing $\varphi{16} = 2$ and so I have just 2 elements of order 2 in $C_{16}$ and so there are just 2 semi direct products. How do I know what these semi-direct products are? Am I right in assuming that one is the trivial semi direct product, i.e the direct product?
2) Another question, I have to say how many semi direct products there are from H to Q (same notation as before. I have 
$H = C_{42}, Q = C_{3}$
So I then get $Aut(H) = C_6 \times C_2$. So from here, I can see that 3 doesn't divide 2 but divides 6 so I just look at that. I want the number of elements of order 3 in $C_6$. I get there are 3 elements of order 3. Assuming $C_6 = \{a, ..., a^6\}$ then the elements of order 3 are $a^2, a^4$ and $a^6$. So from here I get there are three semi direct products for this question. My question is this though, lets assume I had something like $Aut(H) = C_6 \times C_3$, then how would I work this question out? As 3 divides 3 from the other $C_3$ so I can't just ignore it like in the question I had to do. What would I do here? I'm assuming I would have to work out the number of semi direct products in this, and then how would I combine the two?
EDIT: Oh hang on, I need to be able to work out the number of elements in a given order to be able to answer my second question don't I? Dang. Still can't do that :(
 A: About your first question, you want to construct $$G=N\rtimes H$$ wherein $N=\Bbb Z_{17}=\langle n\rangle, H=\Bbb Z_{2}=\langle h\rangle$. You started correctly because we need some homomorphisms which can construct the proper relations for $G$'s. So you took $$\phi:H\to Aut(N)$$ You noted that correctly that $|Aut(N)|=16$ but we should be careful that any generators of $N$ would be mapped to the any generators of $N$. If we define $$\phi_h, h\in H$$ then it is clear that $\phi_h(n)\in N$. In fact, $\phi_h\in Aut(N)$. So there is for example $l$, $0\leq l\leq 16$ such that $$\phi_h(n)=n^l$$ Note that $H,N$ both are cyclic. Since we assume that $\phi_h$ for all $h\in H$ is an automorphism and $|N|=17$ then we should have $(l,17)=1$. In fact we define $$h^{-1}nh=n^l,\;\;0\leq l\leq 16,\;\;(l,17)=1$$
These are well defined relations for our desired $G$.
A: You seem to be seeking semidirect products $\,C_{17}\rtimes C_2\,$ , for which we must know the possible homomorphisms $\,C_2\to\operatorname{Aut}(C_{17})\cong C_{16}\,$ . Since $\,C_{16}\,$ has one single element of order $\,2\,$ there is one unique non-trivial homomorphism as above, which gives the non-abelian group of order $\,34\,$, determined by the action $\,c\cdot a=:a^c=a^{-1}\,\,,\,\,C_2=\langle c\rangle\,\,,\,\,C_{17}=\langle a\rangle \,$ . Thus, we can say that
$$C_{17}\rtimes C_2=\left\{(c^i,a^j)\in C_{17}\times C_2\;\;;\;\;0\leq i\leq 16\,\,,\,0\leq j\leq 1\right\}$$
under the operation 
$$(c^i,a^j)(c^k,a^r):=(c^i(c^k)^{a^j},a^{j+r})$$
The other homomorphism is the trivial one, which gives us the direct product $\,C_{17}\times C_2\cong C_{34}=$the cyclic group of order $\,34\,$
As for your second question: $\,\operatorname{Aut}(C_{42})\cong C_2\times C_6 ,\,$ so how many homomorphisms $\,C_3\to C_2\times C_6\,$ can you find?...
