# Number of Automorphisms on group

I had a doubt in the following question

How many automorphisms possible on the group $Z^{+}_{12}$. Although I believe it should be $12!$ but want to confirm.

Thanks

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Suppose $p$ is a homomorphism that maps 1 to 0. Could $p$ be an automorphism? – MJD Aug 17 '14 at 18:01

Since $\mathbb{Z}_{12}$ is cyclic, an automorphism is determined by what happens to a generator, say $1$, and the generator $1$ must map to a generator of the cyclic group.

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Thanks, but can you pls tell why number of generators is equal to number of automorphisms? – codeomnitrix Aug 19 '14 at 2:52
An automorphism is an isomorphism from a group to itself. Thus an automorphism preserves element orders. In the case of a cyclic group, this means that a generator must map to a generator. So in $\mathbb{Z}_{12}^+$ we could have an automorphism where $1\mapsto 5$, and then the automorphism is completely determined--$x\mapsto 5x\pmod{12}$. The generators of $\mathbb{Z}_{12}^+$ are those elements relatively prime to $12$, which allows you to determine the number of automorphisms. – paw88789 Aug 19 '14 at 10:16

A cyclic group has a single generator, so a homomorphism $\phi$ from a cyclic group to any group is uniquely determined by $\phi$ sends $1 \in \mathbb{Z_{12}^+}$. 5, 7 and 11, being coprime to 12, also share this property of totally determining the homomorphism.

An automorphism is an isomorphism, so the image of $1$ needs to have an order of 12 for our homomorphism to be an isomorphism. Generally, an isomorphism from a cyclic group to itself must send a generator to a generator. So clearly 12 can't be the answer, as sending $1$ to $2$, for example, will produce a homomorphism to $\mathbb{Z_6}$.

Can you finish it from here?

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As do $5, 7$... – amWhy Aug 17 '14 at 18:00
Thanks, but can you pls tell why number of generators is equal to number of automorphisms? Also your last line I am not able to understand it "So clearly 12 can't be the answer, as sending 1 to 2, for example, will produce a homomorphism to Z6." can you please explain it a bit – codeomnitrix Aug 19 '14 at 2:51

Hint:

Think about the generators of $\mathbb Z_{12}$. Indeed, every element which is coprime to $12$ can generate the whole group. $$\varphi(12)=\varphi(3)\varphi(4)=2\times2=4$$

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how you did this $\phi(12)=\phi(3)\phi(4)=2*2=4$ – codeomnitrix Aug 19 '14 at 2:47
@codeomnitrix: By $\phi$, I meant the Euler's phi function in which we can find the numbers which is coprime to 12. – Babak S. Aug 19 '14 at 14:38