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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?

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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 $k$.

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  • $\begingroup$ Could you prove your first statement in the direction of " $k$'s order in the codomain divides 48 implies $\phi_k$ is a homomorphism" $\endgroup$ – MKeller Nov 5 '18 at 7:12
  • $\begingroup$ I edited the answer with more details. If you haven't seen this fact before, I encourage you to prove that if the kernel of a homomorphism contains a normal subgroup of the domain, then the homomorphism factors through the quotient given by the domain over that normal subgroup. $\endgroup$ – Luca Carai Nov 5 '18 at 17:14
  • $\begingroup$ 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 $48$ $\endgroup$ – MKeller Nov 6 '18 at 0:48
  • $\begingroup$ 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 $48$. $\endgroup$ – Luca Carai Nov 6 '18 at 1:31

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