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Here photo of $1$ is $3$ and also $|3| = 10 $ which does not divide $12$, the order of $1$. So this can"t be homomorphism.
Is this correct?
This "map" is doomed from the outcollection, as it is not well defined.
As it fails to be a bonified attribute, it absolutely can"t be a homomorphism.
Besides, it is straightforward to check out the just homomorphism, various other than the trivial one, from $Bbb Z_12$ to $Bbb Z_10$ is $h$, offered by $h(1)=5$.
Yes, your dispute looks excellent.
Call your map $f$. On the one hand also $f(0)= f(12cdot 1)$. Thus, for $f$ to be a homomorphism, one need to also have actually $$ 0 = f( 0 ) = f(12cdot 1) = 12 cdot f(1) .$$So (a variable of) $12$ would certainly need to kill $f(1)$ in $steustatiushistory.orgbb Z_10$ - however $f(1)=3$, and also $$ 12 cdot 3 = 6 ot= 0in steustatiushistory.orgbb Z_10.$$So no element of $12$ kills $f(1)$.
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