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I want to work out $H \rtimes Q$, where $H = C_{17}$ and $Q = C_{2}$. What this means is that I want to work out the groups that map $\theta: C_2 \rightarrow Aut(C_{17})$.

$Aut(C_{17}) \cong C_{16}$. I know that 2 divides 16 and from here I get two SDP's. One is the direct product, $C_{17} \times C_2$ and the other is the SDP I have to find. What I have said is this:

If we denote $H = \langle a | a^{17} = 1 \rangle, Q = \langle b | b^2 = 1 \rangle$ and $Aut(C_{17}) = \langle \mu | \mu^{16} = 1 \rangle$, then I want to find some $\mu \in Aut(C_{17})$ which will map elements of order $2$ in $Aut(H)$ to elements of order $2$ in $H$, i.e some

$$\mu \in Aut(C_{17}) \mathrm{\, such \, that\, } \mu : a \mapsto a^k \mathrm{\,goes \, to\,} \mu^{2} : a \mapsto a^{k^2} $$

In other words, I want the $k$'s such that $k^2 \equiv 1 \mod 17$. Is this right so far? I'm stuck on how I go about trying to answer the question now.

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Hint: Since the automorphism group of $C_{17}$ is cyclic, there is only one automorphism of order $2$. In order to figure out what it is, I will tell you that for any abelian group, there is an automorphism which works in the exact same way as this one. – peoplepower Jan 10 '13 at 0:52
@peoplepower I get that there is one element of order 2, $a^{8}$, but I don't see how this relates to the automorphisms, or does it? Unless that 8 has something to do with my $k$'s? – Kaish Jan 10 '13 at 0:55
Unless its something like: Obviously both $C_{17}$ and $C_2$ are abelian, and as I have worked out one automorphism to be the direct product, is the other one also a type of direct product, i.e multiplication of the eleements? As both these automorphisms work in the samy way – Kaish Jan 10 '13 at 0:57
there is only one nontrivial map $C_2\to \text{Aut}(C_{17})\cong C_{16}$ (since there is only one subgroup of order two). every abelian group has at least one automorphism of order two: inversion. – yoyo Jan 10 '13 at 1:09
@yoyo Yeah I got that, thats what I did in my working out, but I don't get how to work out what this automorphism is – Kaish Jan 10 '13 at 1:12
up vote 1 down vote accepted

So I had assumed previously that you knew what an automorphism group is. Perhaps you need some elaboration on this. I'm going to go through this particular example step by step as thoroughly as I can.

$\text{Aut}(C_{17})$ is a cyclic group of order $16$, but it is not $\mathbb{Z}_{16}$ (just isomorphic to it). It is generated by the bijective homomorphism $\alpha:C_{17}\rightarrow C_{17}$ defined by $\alpha(x)=x^3$. So $\alpha^2$ is the map $x\mapsto x^{3^2}$, $\alpha^3$ is the map $x\mapsto x^{3^3}$, etc. Because of number theory stuff, $\alpha^{16}$ is the map $x\mapsto x$, which is the identity homomorphism $\text{id}$ on $C_{17}$, and no other $\alpha^{i}$ is the identity for $i\leq 15$. So $$\text{Aut}(C_{17})=\{\text{id},\alpha,\alpha^2,\ldots, \alpha^{15}\}.$$ We want to find a homomorphism $\theta:C_2\rightarrow \text{Aut}(C_{17})$ to define the semidirect product. You already found that the homomorphism $\theta_1:C_2\rightarrow \text{Aut}(C_{17})$ defined by $\theta_1(x)=\text{id}$ yields $C_{17}\rtimes_{\theta_1}C_2=C_{17}\times C_2$. This is one of the two possible semidirect products. We find the other one by noticing that the order of $a$ (the generator of $C_2$) is $2$, so it needs to map to the automorphism $\alpha^8$, because $(\alpha^8)^2=\alpha^{16}=\text{id}$. So let $\theta:C_2\rightarrow \text{Aut}(C_{17})$ be defined by $\theta(a)=\alpha^8$.

Remember that $x^3$ means $3x$ in $\mathbb{Z}_{17}$, since the operation is addition. The relation to DonAntonio's comment is that $x^{3^8}$ means $3^8x$ in $\mathbb{Z}_{17}$. But $3^8\equiv 16 \pmod{17}$, i.e. $3^8\equiv -1 \pmod{17}$. So the map $\alpha^8$ is the map $x\mapsto -x$.

You asked in the comments if you should always use $x\mapsto -x$ for an automorphism of order $2$. The answer is no, because it is not true in general that this map is a homomorphism. In fact, it is if and only if the group is abelian. You should prove this real quick. (Hint: Let $\phi:x\mapsto x^{-1}$. $b^{-1}a^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$.) So only with abelian groups can you use this as an automorphism of order $2$ (though it may not be the only automorphism of order $2$).

So what does this mean for your semidirect product $C_{17}\rtimes_{\theta} C_2$?

We proved yesterday that $ba=a\theta(a^{-1})(b)$. In $C_2$ $a=a^{-1}$, so $ba=a\theta(a)(b)$. As we determined above, $\theta$ maps $a$ to the automorphism $x\mapsto x^{-1}$, so $\theta(a)(b)=b^{-1}$, so $ba=ab^{-1}$. We can rewrite that as $b^a=b^{-1}$, yielding the familiar group presentation $$C_{17}\rtimes_{\theta} C_2=\langle a,b|a^2,b^{17},b^a=b^{-1}\rangle$$ which we recognize as the dihedral group $D_{17}$.

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Hi, thank you very much for this. I'm slowly working through it, my first question (and hopefully last) is this: Why is the homomorphism $\alpha :C_{17} \rightarrow C_{17}$ definined by $\alpha(x) = x^3$. Where do you get the cubed from? Why not 2 or 5 or anything else? – Kaish Jan 10 '13 at 23:34
Also, After getting $3^8 \equiv -1 \mod 17$, from here you get that $x \rightarrow -x$. Let's say that you got $3^k \equiv 3 \mod 17$, (or whatever, even if it isn't $\mod 17$), then would your $\theta$ map $\alpha^k: x \rightarrow 3x$? – Kaish Jan 10 '13 at 23:59
We pick $3$ because we want a number whose multiplicative order modulo $17$ is $16$. There is no general way to show you how to determine the generator of $\text{Aut}(C_p)$ (this is an open problem), but in small examples it shouldn't be too hard to just compute $2^i\mod{17}$ until you find that $2^8\equiv 1\pmod{17}$ (meaning that $2$ has multiplicative order $8$, not $16$, and thus is not a generator), then move onto $3$ and compute $3^i\mod{17}$ until you know that the smallest $i$ for which $3^i\equiv 1 \pmod{17}$ is $i=16$. – Alexander Gruber Jan 11 '13 at 0:21
Second question: yes it would. – Alexander Gruber Jan 11 '13 at 0:22
Right, all this theory has just gone out of the window because I made a mistake in my typing. I'm actually mapping $C_7 \rightarrow C_{28}$, so if $\alpha$ generates $C_{28}$, I want to map an element of order 7 to $\alpha$. Order of $\alpha = \varphi(28) = 12$. I now have 6 elements of order 7, $\alpha^{4,8,12,16,18,20,24}$. Now from here, I can't find any $d$ such that $7^d \equiv 1 \mod 28$ as gcd$(7,28) = 7$. What do I do? – Kaish Jan 11 '13 at 1:58

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