Help with what a homomorphism means in $\Phi: (\Bbb{Z}_{18},+_{18})\rightarrow(\Bbb{Z}_{12},+_{12})$

Question

Let $\Phi: (\Bbb{Z}_{18},+_{18})\rightarrow(\Bbb{Z}_{12},+_{12})$ be a homomorphism such that $\Phi([1])=[a]$

Determine values of $a$ for which such a homomorphism exists.

I understand what a homomorphism is, but I do not understand what $\Phi([1])=[a]$ actually means in the context of this problem. What if $a=0$? What does this mean?

Answer 1

Assuming that the notation $[x]$ refers to the equivalence class of $x$, then the notation $\Phi([1])=[a]$ means that under the homomorphism $\Phi:\mathbb Z_{18}\to\mathbb Z_{12}$, the element $[1]$ is mapped to $[a]$. The idea of the problem is that, possibly, a homomorphism can never send $[1]$ to $[a]$ for some $a\in\mathbb Z_{12}$, because this contradicts certain properties of group homomorphisms. The question asks you to determine for which $a$ no contradiction arises and such a homomorphism exists.

Suppose $\Phi([1])=[0]$ (i.e. $a=0$). Then for any integer $n$, we have that $$\Phi([n])=\Phi(\underbrace{[1]+[1]+\dots+[1]}_{n\text{ times}})=\underbrace{\Phi([1])+\Phi([1])+\dots+\Phi([1])}_{n\text{ times}}=n\Phi([1])=n[0]=[0],$$ so that $\Phi$ is the trivial homomorphism, sending all elements to $0$. As another example, in the case $a=1$, we have $\Phi([n])=[n]$ for all $n$ (by the same logic as above). But look at what happens when $n=18$. Inside $\mathbb Z_{18}$, $[18]=[0]$, but we have on one hand $\Phi([0])=[0]$, and $\Phi([18])=[18]=[6]$ on the other, and $[0]\neq[6]$ in $\mathbb Z_{12}$, which is an evident contradiction. Now, try generalising this to look at all $a$s.

Answer 2

The fundamental constraint you have is that because $18 \cdot [1]=[0]$ (where $n \cdot$ represents addition repeated $n$ times), you must have $\Phi(18 \cdot [1])=18 \cdot \Phi([1])=[0]$, so that $o(\Phi([1])|18$.

So which elements of $\Bbb Z_{12}$ have order dividing $18$? Those would be multiples of $2$. Possible values of $a$ are therefore $0, 2, 4, 6, 8, 10$.