Find a homomorphism $f : \mathbb{Z}_{15} \to\mathbb{Z}_5$, and indicate its kernel.

Question

Wondering if this is a good enough answer. Let me know if I did anything wrong. Please, if you have a simpler way let me know.

Since $\mathbb{Z}_{15}$ is cyclic, one can pick ANY generator, $k = 1, 2, 4, 7, 8, 11, 13$, or $14$ and then pick ANY element of $\mathbb{Z}_{5}$ to be the image, $f(k)$. {$\mathbb{Z}_{15}$ is determined by a generator $k$, and the condition: $15k=0$. So the only condition on "$f$" is that $0 = f(0) = f(15k) = 15f(k) = 5(3f(k))$, which is true for all possible $(3f(k))$ in $\mathbb{Z}_{5}$. Hence there are exactly five distinct group homomorphisms.}

Let's take: $f(7)=3$. That completely determines the homomorphism. For example

$f(2)= f(7+7+7+7+7+7+7+7+7+7+7) \\= f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) \\= 3+3+3+3+3+3+3+3+3+3+3 \\= 3$.

To find the kernel, one could just list $f(k)$ for all $k$ in $\mathbb{Z}_{15}$, the kernel is those "$k$" for which $f(k)=0$ For example:

$f(10) = f(7+7+7+7+7+7+7+7+7+7) \\= f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) + f(7) \\= 3+3+3+3+3+3+3+3+3+3 \\= 0$,

so 10 is in the kernel for this "$f$". It turns out that there are only two possible kernels, depending on which group homomorphism you pick.

Alternatively, the kernel must be a subgroup of $\mathbb{Z}_{15}$, for which the order of $\mathbb{Z}_{15}$ divided by the order of that subgroup produces a divisor of the order of $\mathbb{Z}_{5}$ {$1$ or $5$}. So the kernel is a subgroup of $\mathbb{Z}_{15}$ of order either $15/5=3$ or $15/1=15$, and there are just one of each. Both are kernels for some homomorphism. However, if $\mathbb{Z}_{15}$ is the kernel then $f(k)=0$ for all $k\in\mathbb{Z}_{15}$, the trivial homomorphism. So for all the non-trivial homomorphisms, the kernel is just the subgroup generated by the elements of order $3$.

Answer 1

Not really an answer, but when I do these kind of problems with homomorphisms, I usually start by seeing where 1 goes. What is $f(1)$ equal to?

To answer your question, your answer is good enough. You made several good conclusions and have the algebra to prove it. You should tidy it up though.

Answer 2

One method is to see that you have cyclic group of order 5, whih has no proper, non trivial subgroups. Since the image of a homomorphism has to be a subgroup, you will know that image must be either identity element, or the whole group. After that it's easy to see kernel must be subgroup of order 3, for onto homomorphism. This makes things easier, when you have a group with no proper, non trivial subgroups.