Finding Kernel for Group and checking homomorphism

Question

Let function f: Z10 -> { 0, 2, 4, 6, 8} (additive group) be defined as f(a)= 2a. How can I check if f is homomorphism? And how to compute 5 + Ker(f) + 2 + Ker(f).

After going through some books and lecture notes I have not find anything useful. I know how to find kernel in others notations, but this one makes me very confused. Any help would be great.

So I need to have for every element to be satisfied f(a+b) = f(a) + f(b). f(0+2) = 2*2 = 4 = f(0) + F(2) = 0 + 4 = 4, so this one is OK! f(2+4) = 2*6 = 12 = f(2) + f(4) = 4 + 8 = 12 OK! f(4+6) = 20 = 8 + 12 =20 so this one also f(6+8) = 28 = 12 + 16 = 28 so now I have homomorphism proved?

At this point I can compute Ker (f). For Kernel I take that element form set which is neutral for addition, so it is 0. Than compute 5 + Ker(f) + 2 + Ker(f) = 7.

NB. If there was not 0 in set there is now way to take Kernel from that set!?

Answer 1

So, I guess, you have $\Bbb Z_5$ there as the range.

It's straight forward to check that you have a homomorphism.

The kernel would be $\{0,5\}$.

The sum of two cosets can be gotten by adding the representatives. So $(5+H)+(2+H)=7+H=2+H$, where $H=\operatorname{ker}f$.

Note that your notation for $\Bbb Z_5$ is a little weird. Normally you'll see $\Bbb Z_5=\{0,1,2,3,4\}$. Note also that $5+H=H$, since $5\in H$.

Answer 2

Assuming that $f$ maps $\mathbb{Z}_{10}$ into the subgroup $\{0,2,4,6,8\}$ of $\mathbb{Z}_{10}$, then, there is a very much more straightforward way to check that $f$ is an homomorphism than yours. Check this:

For any $a$ and $b$ in $\mathbb{Z}_{10}$ we have $f(a+b) = 2(a+b) = 2a+2b = f(a)+f(b)$.

Also, your solution is not entirely correct because you are missing cases as, for example, $f(1+2) = f(1)+f(2)$. You see?

Also, the kernel of $f$ is not "zero". The kernel of $f$, denoted $\ker f$, is the set of all $a \in \mathbb{Z}_{10}$ such that $f(a) = 0$, that is, $2a=0$. Of course, the elements that satisfies precisely this are $0$ and $5$, so, $\ker f = \{0,5\}$.

Now, when we write $a+H$, for $a \in \mathbb{Z}_{10}$ and $H$ a subgroup of $\mathbb{Z}_{10}$, $a+H$ is the set of all elements of the form $a+b$ for some $b \in H$. Then, for example, $$7 + \ker f = 7 + \{0,5\} = \{7,12\} = \{2,7\}.$$

Can you finish this?