A well defined homomorphism from $\mathbb{Z}/48\mathbb{Z}$ into $\mathbb{Z}/{36}$ I read book of Dummit and Foot Abstract algebra. I need some help with the following question.

Let $\mathbb{Z}/{36} = <x>.$ For which integers $a$  does the map $\psi_{a}$ defined by $\psi_{a}: \bar{1} \mapsto x^a$ extend 
  to a well defined homomorphism from $\mathbb{Z}/48\mathbb{Z}$ into $\mathbb{Z}/36\mathbb{Z}$ Can $\psi_{a}$ ever be a surjective 
  homomorphism? 

Here are some attempts.
Let $n+m<48.$ We have $$\psi_a (\bar{n}+\bar{m})=x^{a(n+m) \mod 36}.$$ On the other hand 
$$\psi_a (\bar{n})\psi_a (\bar{m})=x^{an\mod 36}x^{am\mod 36}=x^{an\mod 36+am\mod 36}.$$
Suppose that $\psi_{a}$ is  a surjective. Then for all $x^i, i=0\ldots 35$ there is $\bar(j), j=0\ldots48.$ such that $\psi_a (\bar{j})=x^i.$ For which values a???
This is my attempt so far.- 
 A: See that $[1]$ is a generator of $\mathbb Z_{48}$ hence $\phi([1])=x^a$ must be a generator of $\mathbb Z_{36}$ which is true if $\gcd(a,36)=1$
Now find the possible values of $a$
By Isomorphism Theorem;
$\dfrac{\mathbb Z_{48}}{ker f}\cong Im f$
For onto we must have o(Im f)=36; is it possible?
A: I am using the additive groups $\mathbb{Z}/m \mathbb{Z}$ for both the domain and range.
First suppose $\psi_a$ is a well defined homomorphism.
We have $\psi_a([1]_{48}]) = [a]_{36}$ and if $\psi_a$ is a homomorphism then
$\psi_a([n]_{48}]) = [an]_{36}$ and so
$\psi_a([48]_{48}]) = [0]_{36} = [48a]_{36}$ from which we get $36 \mid 48a$. This
gives $3 \mid 4a$ and hence $3 \mid a$.
Now suppose $3 \mid a$, then we first need to check that $\psi_a$ is well defined in the sense that if $[x]_{48}=[y]_{48}$ then $\psi_a([x]_{48})=[ax]_{36} = \psi_a([y]_{48})=[ay]_{36}$,
or equivalently, $[a(x-y)]_{36}=[0]_{36}$. Since $x-y = 48 k$ for some $k$ we see that $48 ak = 12 \cdot 4 \cdot a k = 4 \cdot 12 \cdot 3 \cdot {a \over 3} k = 36 \cdot 4 \cdot {a \over 3} k$ and so $[a(x-y)]_{36}=[0]_{36}$.
Then since
$\psi_a([n+m]_{48}]) = [a(n+m)]_{36} = [an]_{36} + [am]_{36} = \psi_a([n]_{48}]) + \psi_a([m]_{48}])$, we see that $\psi_a$ is a homomorphism.
Note that $\operatorname{im} \phi_a = < [a]_{36}>$, and Proposition 5 (2) in Dummit & Foote shows that
$|\operatorname{im} \phi_a| = {36 \over \gcd(a, 36)} $. Since $3 \mid a$, we see
that $\gcd(a, 36) \ge 3$ and so $|\operatorname{im} \phi_a| \le 12$ and so $\psi_a$
cannot be surjective.
