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I'm trying to construct a homomorphism $\theta:C_2 \rightarrow AutC_{17}$. To do this I need to map the generator of $C_2$ (call it $a$) to a generator of $AutC_{17}$, but to do this I need a way of finding generators of an automorphism group.

Does anybody know a method of doing this? Many thanks.

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$C_{n}$ is the cyclic group of order $n$, i.e. $C_{3}={(a, a^2, a^3 =1)}$ Also I probably should have mentioned that I'm looking for non-trivial homomorphisms, so excluding mapping $a$ to the identity. Thanks for your replies so far. –  Traxter Nov 2 '12 at 0:12

2 Answers 2

up vote 3 down vote accepted

The automorphism group of $C_{17}$ is isomorphic to $C_{16}$. If you want to embed $C_2\cong\langle y \rangle$ into it, then letting $C_{16}\cong \langle x \rangle$, you can just send $y\mapsto x^8$. Of course, this isn't very satisfying, and you probably want to actually construct the automorphism itself; this requires some explanation.

If you've taken number theory, you've seen $(\mathbb{Z}/17\mathbb{Z})^\times$, the group of units modulo $17$. If you've run into finite fields, you know that the additive group of $\mathbb{F}_{17}$ is isomorphic to $\mathbb{Z}/17\mathbb{Z}$ and the multiplicative group of $\mathbb{F}_{17}$ is $(\mathbb{Z}/17\mathbb{Z})^\times$. If you haven't seen either of those yet, it turns out that if you multiply the integers in a cyclic group of prime order rather than adding them, you get a group as long as you remove $0$ (since it has no multiplicative inverse). You can read about this in any elementary text on modular arithmetic (including Wikipedia).

What I am getting at is that there is a rather elegant description of the automorphism group of a cyclic group by multiplications of its elements. If we take an $a\not= 0\in \mathbb{Z}/17\mathbb{Z}$, we can define the function $\theta_a:\mathbb{Z}/17\mathbb{Z}\rightarrow \mathbb{Z}/17\mathbb{Z}$ by $\theta_a(x)=ax$. Noting that $$\theta_a(x+y)=a(x+y)=ax+ay=\theta_a(x)+\theta_a(y)$$ and that $\theta_a(x)=ax=0$ implies that $x=0$, we see that this is an automorphism of $\mathbb{Z}/17\mathbb{Z}$.

We can show that such automorphisms form a group under composition. $$(\theta_a\theta_b)(x)=\theta_a(\theta_b(x))=a(bx)=(ab)x=(\theta_{ab})(x)$$ so $\theta_a\theta_b=\theta_{ab}$. (Here, you can see the isomorphism to $(\mathbb{Z}/17\mathbb{Z})^\times$ popping up in the subscript: the map $\theta_a\mapsto a$.) The rest of the group axioms are verified easily. It turns out that this group of $\theta_a$'s constitutes the entire automorphism group of $\mathbb{Z}/17\mathbb{Z}$.

So, finding a nontrivial embedding of $C_2$ into $C_{17}$ reduces to finding an element $a\in (\mathbb{Z}/17\mathbb{Z})^\times$ with order $2$. Noting that $16\equiv-1\mod{17}$, you see that the map is $\phi:C_2\rightarrow \text{Aut}{C_2}$ by $\phi:y\rightarrow \theta_{16}$. Of course, this method will work to find the automorphism group of $C_p$ for any prime $p$, and the automorphism group of any cyclic group $C_n$ can be described similarly as long as the $a$'s in $\theta_a$ correspond to units modulo $n$. (For a good time, try forming the semidirect product of $C_n$ by $\text{Aut}{C_n}$ - you now know how to make a holomorph.)

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Wow. I really didn't expect such an in depth explanation but I understand it now, thank you so much for your help! –  Traxter Nov 2 '12 at 7:16

You seem to be trying some groups semi-direct stuff. Now, in order to get a non-trivial homomorphism as you want you must map the generator of $\,C_2\,$ to an element of the same order in $\,\operatorname{Aut}(C_{17})\cong C_{16}\,$ .

If $\,C_{17}=\langle\,b\,\rangle\,\,,\,\,C_2=\langle\,a\,\rangle\,$ ,then you can always try the nicest, cutest automorphism of an abelian group: the involution $\,x\to x^{-1}\,$ , thus:


If you're really going on semidirect products, you can write this as $\,b^a:=b^{-1}\,$

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@Traxter: You could also take $$f:C_2\to\text{Aut}(C_{17})\\ x\longmapsto f_x\in \text{Aut}(C_{17})$$ with $f_x(y)=y^x=y^l$ and $(l,17)=1$ and $l^2\equiv 1\; \; \text{mode 2}$. +1 for really the cutest map. –  Babak S. Nov 2 '12 at 2:20

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