Question on finding automorphism group of a group.

Question

While reading a book "contemporary abstract algebra by J.A. Gallian" i saw in Example 13 of chapter 6 (isomorphism) that,

to find find $Aut(\mathbb{Z}_{10})$ we determine choices for $\alpha (1)$. Clearly there are four possibilities for $\alpha (1)$ they are $\alpha (1)=1$, $\alpha (1)=3$, $\alpha (1)=7$, $\alpha (1)=9$. After this author denotes above mappings respectively by $\alpha_1$,$\alpha_3$,$\alpha_7$,$\alpha_9$. Then author shows $\alpha_3$ is onto. Upto this i understand all.

But after this, it is written that $\alpha_3(a+b)=3(a+b)= 3a+3b=\alpha_3(a)+\alpha_3(b)$

and hence $\alpha_3$ is operation preserving.

I didn't get that how $\alpha_3(a+b)= 3(a+b)$? (Since while writing this isn't author already used the fact $\alpha_3$ is operation preserving?)

I know $\alpha(k)=\alpha(1+1+...+1(\text{k terms}))=\alpha(1)+\alpha(1)+...+\alpha(1)=k\alpha(1)$

this is because $\alpha$ is homomorphism (operation preserving) but in case of $\alpha_3$ we have to prove that it is homomorphism (operation preserving) so how can we directly say $\alpha_3(a+b)= 3(a+b)$? Please help.

Answer 1

Probably the intended argument is something like this:

If $f$ is operation preserving, then it has to be multiplication by $f(1)$, since as you say we can write

$$f(k)=f(1+1+1+\ldots+1) = f(1) + \ldots + f(1) = k f(1)$$

That is, this is a necessary, but maybe not sufficient, condition for $f$ to be operation preserving.

So now we need to check whether such a map actually is operation preserving for all pairs of summands, since a priori we only handled the case where one of the summands is 1.

Answer 2

An automorphism is determined by what it does to a generator, in this case $1$, of $\Bbb Z_{10}$. Since we are given that $\alpha_3(1)=3$, it follows that for any $g\in \Bbb Z_{10}$, that $\alpha_3(g)=g\cdot\alpha_3(1)=g\cdot3=3g$. That $\alpha_3$ is operation preserving was used, but that is ok because $\alpha_3$ was assumed to be an automorphism.