Determine the number of surjective Homomorphism.

Question

Good evening,

I'd like to discuss the following exercise with you :

Let $G=\mathbb{Z}/18\mathbb{Z}\times \mathbb{Z}/12\mathbb{Z}$ and $G' =\mathbb{Z}/36\mathbb{Z}$.

The second point of the exercise asked to count the number of surjective Homomorphism $f : G \to G'$.

In those cases, I usually consider that a homomorphism to be surjective must have the image of the generator, in this case $[1]_{36\mathbb{Z}}$.

In this case the solution gives a condition that i don't understand, it says :

"Suppose that $f(1,0)=\bar r$ and $f(0,1)=\bar s$.

Necessarly $\bar r=2\bar r_1$ and $\bar s=3\bar s_1$, because the orders of $\bar r$ and $\bar s$ have to divide 18 and 12.

The Homomorphism is defined by $f(x,y) = rx+sy$, and it is surjective if and only if exist $x,y$ such that $\bar rx+\bar sy$ be a generator of $\mathbb{Z}/36\mathbb{Z}$ , so if exist $x, y$ such that $(2r_1x+3s_1y, 36)=1"$.

I really don't understand the condition, an explicative answer would be appreciated,

Thanks.

Answer 1

Hint: To approach the question from another angle, note that $f$ is surjective iff there is $u \in G$ such that $f(u)$ has order $36$. This can only happen if the order of $u$ is a multiple of $36$. Now there are no elements of order $72$ in $G$, and so $u$ must have order exactly $36$. So the question reduces to: how many elements of order $36$ are there in $G$?

Answer 2

If a homomorphism is surjective, then its range contains a generator of $\Bbb Z/36\Bbb Z$, like $\bar 1$ for instance (in fact it contains all of them).

Conversely, if the range of a homomorphism contains a generator of $\Bbb Z/36\Bbb Z$, then the homomorphism is surjective (very easy exercise).

Finally, generators $\Bbb Z/36\Bbb Z$ are exactly those $\bar k$ such that $\gcd(k,36)=1$.