Take $\mathbb{Z}_{15}$ and $\mathbb{Z}_{18}$, and find all the group homomorphisms. I'm having trouble with the concept, so the specific example is not necessarily important.

From the definition of a group homomorphism the only restriction is that the homomorphism has to satisfy the property $\phi(x + y) = \phi(x) + \phi(y)$.

Now since $\mathbb{Z}_{15}$ is cyclic with $1 \in \mathbb{Z}_{15}$ as the generator, any element $x \in \mathbb{Z}_{15} = 1+1+\cdots+1$ ($x$ times), so $\phi(x) = \phi(1+1+\cdots+1) = x\phi(1)$.

Now on another post this same example came up, and it was noted that the map defined by the relation $\phi(1) = 2$ is not a homomorphism. But why not?

Since $\phi(x + y) = 2(x + y) = 2x + 2y = \phi(x) + \phi(y)$, why isn't this a homomorphism?

Please any help, I know I'm missing something important.

  • 1
    $\begingroup$ Do you mean group homomorphisms $\phi \colon \mathbb Z_{15} \to \mathbb Z_{18}$? We know the identity maps to the identity, so $\phi(0) = 0$. Now if $\phi(1) = \alpha$, then $\phi(n) = n\alpha$, so in particular, $\phi(0) = \phi(15) = 15\alpha$; thus if non-trivial homomorphisms exist, then $15\alpha \equiv 0 \pmod {18}$. There may be some case work to do from here, but this limits the possibilities for $\alpha$. It may not be the case that all the $\alpha$ work out; you must check these yourself. $\endgroup$ – user217285 May 11 '16 at 17:29

You are missing that you are working with classes. For example in $Z_{15}$ if you state that $\phi(1)=2$ then you have that $\phi(10+6)=\phi(1)=2$ and $\phi(10)+\phi(6) = \overline{32} = 14\in Z_{18}$ so it is not an homomorphism of groups.

  • $\begingroup$ Thank you! Now it makes sense. $\endgroup$ – Lee Nguyen May 11 '16 at 17:50

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