How to determine when this is a well-defined homomorphism between cyclic groups I've been having trouble with this concept. I'm not very familiar with cyclic group structure. Question: Let $Z_{36} = \langle x \rangle$. For which integers $a$ does the map $\phi_a$ defined by $\phi_a : \bar{1} \to x^a$ extend to a well-defined homomorphism from $\mathbb{Z}/48\mathbb{Z}$ to $Z_{36}$? I know firstly that $\phi(a \star b) = \phi(a) * \phi(b)$ must hold for all $a,b \in \mathbb{Z}/48\mathbb{Z}$.
Also, is the domain of $\phi_a$ all of $\mathbb{Z}/48\mathbb{Z}$? I'm not sure what's going on. Any help appreciated.
 A: let $f:G\rightarrow H$  a group morphism and $G / K$
 a quotient group , then
$f$ can be factored through the  quotient group $G/K$ iff
$kerf\supseteq   K$.
Let $\phi_a:\Bbb{Z}\rightarrow \Bbb{Z}/n\Bbb{Z}$ the morphism that assign $1$ to
$a{\bar1}$  (in additive notation) so $\phi_a$ can factorized via
the quotient $\Bbb{Z}/m\Bbb{Z}$ iff $m\Bbb{Z}\subseteq ker\phi_a$. but
$ker\phi_a=\{r\in \Bbb{Z} \mid 36 $ divide $ar \}$ that is iff for some
integer  $k$ ,  $r=\frac{36}{gcd(a,36)}k $, so
$ker\phi_a=\frac{36}{gcd(a,36)}\Bbb{Z}$  and  so   for all
 integer  $a$ with $3$ divide $a$ we have   $48\Bbb{Z}\subseteq \frac{36}{gcd(a,36)}\Bbb{Z}$ and  $\phi_a$ can extended to $\Bbb{Z}/48\Bbb{Z}$
A: First, given that neither 36 nor 48 is prime, these must be the additive cyclic groups (since every element of the group must have an inverse) so it might simplify things to write this out additively as:
$$\phi_a:\mathbb{Z}/48\mathbb{Z}\rightarrow\mathbb{Z}/36\mathbb{Z}$$ defined by 
$$\phi_a(1)=ax$$ and the homomorphism quality that $$\phi_a(b+c)=\phi_a(b)+\phi_a(c)$$ as opposed to writing it generally with $*$ and $\star$
Now it seems much more clear that given $\phi_a$ is a homomorphism, we've got:
$$\phi_a(\alpha)=\phi_a(1+...+1)=\phi_a(1)+...+\phi_a(1)=\alpha\phi_a(1)$$
The question now becomes: For which values of $a\in\mathbb{Z}$ does the homomorphism quality hold? Assuming it does hold we see:
$$\phi_a(b+c)=\phi_a(b)+\phi_a(c)=b\phi_a(1)+c\phi_a(1)=b(ax)+c(ax)=(b+c)(ax)$$
which is true for $\forall a \in \mathbb{Z}$. The domain of any function must be the entire set, so yes, the domain is all of $\mathbb{Z}/48\mathbb{Z}$. It's been a while since I've done much group theory, so somebody else should certainly confirm that what I've written is accurate, but as far as I'm concerned, $a$ could be any integer. Cheers!
