So I am trying to prove $f([a]+[b]) = f([a]) + f([b])$ for $\mathbb Z$ mod $12$ and the same for multiplication... Can you show that this is true based on a function being surjection and injective and any combination of the two without showing every possibility of every element in $\mathbb Z$ mod $12$? Is there any definitions stating this?
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Where does $f$ map from/to? – Berci Apr 01 '14 at 20:05
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Zmod 12 to Zmod4 for the function f(x)=3x – D-Man Apr 01 '14 at 20:07
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Exactly how the original exercise was worded?? – Berci Apr 01 '14 at 20:08
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it says it is mapped from Zmod12 to Zmod4 defined by f(x)=3x – D-Man Apr 01 '14 at 20:09
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So, if I understand right, you are given $f(x)=3x$, defined on $\Bbb Z_{12}$.
This maps $0\mapsto 0,\ 1\mapsto 3,\ 2\mapsto 6,\ 3\mapsto 9,\ 4\mapsto 0,\ \dots$.
We can say that $f:\Bbb Z_{12}\to\Bbb Z_{12}$, in this case it is neither injective nor surjective. (Why?)
The range of $f$ is $im(f)=\{0,3,6,9\}\subseteq\Bbb Z_{12}$ which is isomorphic to $\Bbb Z_4=\{0,1,2,3\}$.
So, we can also write $f:\Bbb Z_{12}\to\, im(f)\ \cong \Bbb Z_4$, and in this sense $f$ is going to be surjective.
And, finally, the fact that $f$ preserves addition follows simply by linearity: $$f(a+b)\ =\ 3(a+b)\ =\ 3a\ +\ 3b\ =\ f(a)\,+\,f(b)\,.$$
Berci
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so is it directly related to being surjective? is there a statement that directly correlates the two? – D-Man Apr 01 '14 at 20:25
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$f$ is not going to be injective if defined on $\Bbb Z_{12}$ as $f(0)=f(4)$. – Berci Apr 01 '14 at 20:30
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yeah i already know this function is surjective and not injective.... i was just asked to show this question for both addition and multiplication. I could list all the elements out in Zmod12 but I just wanted a good fact to prove this easily because it was surjective and not injective. ... I think addition holds just not multiplication – D-Man Apr 01 '14 at 20:33
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