# Finding Kernel for Group and checking homomorphism

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!?

• The Kernel of a group are the elements which will be mapped to the 0 element. What are those elements?
– Paul
Commented May 6, 2020 at 22:35
• Commented May 6, 2020 at 22:36
• To find the kernel you must first check that the map is a morphism. Right now, what you have does not even begin to make sense, because the set $\{0,2,4,6,8\}$ is not a group under the standard addition, so you need to be a lot more careful about what it is you are doing. Once you know you have a group, check whether $f(a+b) = f(a)+f(b)$ holds for all $a,b\in\mathbb{Z}_{10}$. If it does, voila! it’s a homomorphism, and then you can find the kernel the usual way: figure out what maps to the identity. Commented May 6, 2020 at 22:38
• @Paul Some language barier, by o element you think neutral element? If so then Ker (f) is 0. Commented May 6, 2020 at 22:38
• There may be context that you aren’t passing along. But without that context, either because the book is missing it or because you haven’t realized what that context is, then the problem, as written, is nonsense. There are some obvious ways to insert meaning to make it reasonable, but that meaning needs to be there. Commented May 6, 2020 at 23:04

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?

• Yes I think I can see it! It was very strange set (every second number) so I thought I have and can to operate and check only with those given numbers, also, by mistake I forgot 5. Commented May 6, 2020 at 23:20

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 $$5\in H$$.

• First of all thanks a lot! I have made some changes in original post, done some computation. Can you take a look at it please Commented May 6, 2020 at 22:59
• Sure. You need to check for all pairs of elements whether the homomorphism property is satisfied. You can do it all at once, by writing $f(a+b)=2(a+b)=2a+2b=f(a)+f(b)$. Good luck getting used to some of these new ideas. As I said, I think you're talking about $\Bbb Z_5$, because there's only one five element group.
– user403337
Commented May 6, 2020 at 23:14