ϕ:Z12→Z41→ϕ(1)a→ϕ(a)=aϕ(1)
It is sufficient to find the value of ϕ(1).
Now o(ϕ(1)) divides 4 and 12. [ o(ϕ(1))∣o(1) and o(1)=12]
Therefore o(ϕ(1))=1or2or4
Now when o(ϕ(1))=1⇒ϕ(1)=1 ,
when o(ϕ(1))=2⇒ϕ(1)=2 ,
when o(ϕ(1))=4⇒ϕ(1)=3
Therefore the possible homomorphisms are a→a , a→2a,a→3a .
So the nontrivial homomorphisms are a→2a and a→3a
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