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I'm trying to find the number of automorphisms of Z8. When I google around, I find stuff like:

There are 4 since 1 can be carried into any of the 4 generators.

The problem hint tells me to make use of the fact that, if G is a cyclic group with generator a and f: G-->G' is an isomorphism, we know that f(x) is completely determined by f(a).

Thing is, I can think of 7! 1-1 and onto mappings of Z8 onto itself. I guess I don't see exactly why 1 has to get carried into a generator...why can't I have f(n) = n + 1 (mod 8), just shifting each element one to the right?

Thanks for any guidance on this, Mariogs

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  • $\begingroup$ Do you know what an automorphism in group theory is? Look it up! When proving things about concepts, it is very important to know and understand definitions! Not just any map on the underlying set is an automorphism. $\endgroup$ – anon Feb 19 '14 at 16:27
  • $\begingroup$ In any case why do you say $7!$ ? There are $8!$ such mappings. $\endgroup$ – Derek Holt Feb 19 '14 at 16:58
  • $\begingroup$ @DerekHolt It looks like Mariogs understood that $0$ maps to $0$ in an automorphism. $\endgroup$ – John Habert Feb 19 '14 at 17:53
  • $\begingroup$ But his example $f(n)=n+1$ does not have that property, so it looks like he didn't understand that at all! $\endgroup$ – Derek Holt Feb 19 '14 at 18:16
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An automorphism is not just one-to-one and onto. It is also a group homomorphism. While there are many one-to-one and onto maps between a set and itself, not all of such maps will preserve the structure of the group. To be a group homomorphism, you must have $f(a+b) = f(a)+f(b)$ for any $a,b\in\mathbb{Z}_8$. Under your shift right suggestion we would get that $f(1+1) = f(2) = 3$ but $f(1)+f(1) = 2+2 = 4$. So it is not a group homomorphism.

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  • $\begingroup$ Ah, I see. So why does this imply that we must have f(1) = one of the generators of Z8 (1, 3, 5, 7)? $\endgroup$ – bclayman Feb 19 '14 at 18:02
  • $\begingroup$ It is the only way to preserve the structure of the group. Since we can define any element of $\mathbb{Z}_8$ as a multiple (or power) of a generator, if we do not take a generator to a generator then we break the group structure. $\endgroup$ – John Habert Feb 19 '14 at 18:08
  • $\begingroup$ Ahhh, that's perfect. Makes a ton of sense, thanks! $\endgroup$ – bclayman Feb 19 '14 at 18:11
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Note that $1$ generates $\mathbb{Z}_{8}$ as a group, so any group morphism $\varphi:\mathbb{Z}_8\rightarrow\mathbb{Z}_8$ is determined by $\varphi(1)$. Furthermore, if $\varphi$ is an automorphism, then $\varphi(1)$ generates $\mathbb{Z}_8$. The possible generators of $\mathbb{Z}_8$ are $1,3,5,7$. It then remains to check that for each possible choice of generator, there exists $\varphi$ with $\varphi(1)$ equal to the generator. This is the case, so there are $4$ possible automorphisms of $\mathbb{Z}_8$. (To see this, define $\varphi(n)=3n, 5n, 7n$ and check each of these.)

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