# Homomorphisms from $\mathbb{Z}/48\mathbb{Z}$ into $\mathbb{Z}/36\mathbb{Z}$

So, this problem is sourced from §2.3.9 in Dummit and Foote's 'Abstract Algebra.' It is: For what values of $$k$$ is $$\phi_k : \mathbb{Z}/48\mathbb{Z} \to \mathbb{Z}/36\mathbb{Z}, \phi (1) = k$$ a homomorphism? My attempt so far:

We need to only consider $$k$$ from $$0$$ to $$35$$, since $$\phi _k$$ is going to be the same map as $$\phi_{k \pmod{36}}$$. Next, I was thinking about how the image of $$\phi$$ will define a subgroup in the codomain. So, the image's order, $$|Im(\phi)|$$ has to divide $$36$$. I also know that the kernel of $$\phi$$ give a subgroup in the domain, so $$|Ker(\phi)|$$ divides $$48$$. I wanted to know if there was a relationship between the sizes of these two.

The book, when proving Lagrange's Theorem, uses that a group action's orbits are all the same size. I felt like there is some relationship here, but I'm not really sure how to articulate it. I thought that fiber over each element in the image should have the same size and partition the domain. So, I proved this and got that $$|Ker(\phi)||Im(\phi)|=|domain|$$.

Next, I thought about how $$k$$ determines $$|Im(\phi)|$$. The image is certainly $$\langle k\rangle$$, and the book gives as a theorem the order of this subgroup is $$\frac{36}{gcd(36,k)}$$.

So now I have two necessary conditions for $$k$$: $$\frac{36}{gcd(36,k)}$$ divides both 36 and 48. Now, I know close to nothing about number theory (I know many important ideas in modern algebra are motivated by number theory, but I really haven't found a book for that I can get excited about).

How can I take these two pieces of information into a way to find possible value for $$k$$ more easily? Moreover, how do I find a sufficient condition, so I don't have to test each $$k$$ that satisfies these individually?

It turns out that $$\phi_k$$ is a homomorphism iff the order of $$k$$ in $$\mathbb{Z}/36 \mathbb{Z}$$ divides 48. Indeed, $$0=\phi(0)=\phi(48)=48k$$. On the other hand, if the order of $$k$$ divides 48, the kernel of the homomorphism $$\psi:\mathbb{Z} \to \mathbb{Z}/36 \mathbb{Z}$$ sending $$1$$ to $$k$$ contains $$48 \mathbb{Z}$$. Therefore, that homomorphism factors through $$\mathbb{Z}/48 \mathbb{Z}$$. That is, there is a homomorphism $$\mathbb{Z}/48 \mathbb{Z} \to \mathbb{Z}/36 \mathbb{Z}$$ mapping $$1$$ to $$k$$.
Thus, a sufficient and necessary condition for $$\phi_k$$ to be a homomorphism is that $$36$$ divides $$48k$$. (why?)
That holds exactly when $$3$$ divides $$4k$$. Since $$3$$ is prime, $$3$$ has to divide $$k$$. Thus the condition is equivalent to $$3$$ divides $$k$$.
• Could you prove your first statement in the direction of " $k$'s order in the codomain divides 48 implies $\phi_k$ is a homomorphism" – MKeller Nov 5 '18 at 7:12
• Thanks. This problem occurs before the discussion of normal subgroups or quotients, so I was trying to avoid using them. I instead proved a little weaker statement, showing that if $|k|$ divides $48$, the homomorphism is well defined. That is, a situation like with $k=1$ where $\phi(36)=0$ but $\phi(36^2)=\phi(24) = 24 \neq 0^2 = 0$ cannot happen if $|k|$ divides $48$ – MKeller Nov 6 '18 at 0:48
• You can prove that the map is well-defined using modular arithmetic without talking about quotients or normal subgroups. $\phi_k$ is well-defined iff for any $x,y$, if $48$ divides $x-y$ then $36$ divides $k(x-y)$. But this is equivalent to $36$ dividing $48k$ because $x-y$ is any multiple of $48$. – Luca Carai Nov 6 '18 at 1:31