I have to find the number of isomorphisms from the group $( \mathbb{Z}_3, + )$ to itself.
I don't know of any procedure to do this so I basically just tried to guess functions until I cannot find anything else. I found that the functions:
$$f(x) = x$$
$$f(x) = \hat{2}x$$
are both bijective and they hold the equality:
$$f(x + y) = f(x) + f(y)$$
true for any $x, y \in \mathbb{Z}_3$. So I concluded that these are all the isomorphism from the group $(\mathbb{Z}_3, +)$ to itself, since I couldn't find any more functions to be bijective and satisfy that condition.
I checked the answer of the exercise and it agrees with me, saying that the correct answer is $2$ (meaning $2$ isomorphisms, I guess the ones I found).
My question is this: Is there a more organized, general way of finding the answer to this question.
Guessing all the possible function seems a bit weird, there are infinite possibilities.
How do I know that I found the maximum number of isomorphisms and that I can stop? Is there a better strategy than guessing the functions and coming up with the answer?
Best Answer
There aren't an infinite number, for it to be a bijection, you have $3$ choices for $f(0)$, $2$ choices for $f(1)$, and then $f(2)$ is determined.
That is, your choices are
$f(0), f(1), f(2) = 0,1,2$
$f(0), f(1), f(2) = 0,2,1$
$f(0), f(1), f(2) = 1,0,2$
$f(0), f(1), f(2) = 1,2,0$
$f(0), f(1), f(2) = 2,0,1$
$f(0), f(1), f(2) = 2,1,0$
and then not all of these will respect the group operation, e.g. for $f(0), f(1), f(2) = 2,1,0$, you have $f(0 + 1) = f(1) = 2$, but $f(0) + f(1) = 2 + 1 = 0$.
This is still a little brute force, but by observing, e.g. that $f(0) = 0$ is true for any group isomorphism, you narrow things down even more.