Question

Need help putting the logic in neat. :) Let G, H,K be groups
f1: G--> H
f2: G--> K
f3: G--> H x K
f3 sends g to (f1(g) , f2(g))
f1, f2 are homomorphisms Prove
f3 is homomorphisms. I have it done
then prove kernel f is \(kernel ~f1\cap kernel~f2\)
Please, help

Answer 1

It doesn't allow me post the long comment, ha

Answer 2

Kernel doesn't make sense unless you have more structure. Are they groups?

Answer 3

yes, they are

Answer 4

You should state that in your question next time.

Answer 5

:) I think people can help me in this site know what I am doing. hihihi... Since not many of helpers here help me

Answer 6

Do you know that an isomorphism has a trivial kernel?

Answer 7

\[kernel f_1=\{g\in G |~f_1(g) = e_H\}\]

Answer 8

oh, I am so sorry

Answer 9

Since it is an isomorphism the kernel contains only one element, the identity element of G.

Answer 10

They are not iso, they are just homomorphism

Answer 11

Can you edit your question to correct that?

Answer 12

\[kernelf_2=\{g\in G| f_2(g) =e_K\}\]

Answer 13

In your question is states: "f1, f2 are isomorphisms,"

Answer 14

I did

Answer 15

This stupid computer limits my typing

Answer 16

\[kernel~f_1\cap kernel~f_2 =\{g\in G ~| ~f_1(g)=e_H \cap f_2(g) = e_K\}\]

Answer 17

So let \(a, b \in G\) we need to show that \(f_3(a \cdot b) = f_3(a)\cdot f_3(b)\). So let us verify.
$$\begin{eqnarray*}
f_3(a\cdot b) &=& (f_1(a \cdot b), f_2(a \cdot b)) \\
&=& (f_1(a)\cdot f_1(b), f_2(a)\cdot f_2(b)) \\
&=& (f_1(a), f_2(a)) \cdot (f_1(b), f_2(b)) \\
&=& f_3(a)\cdot f_3(b)
\end{eqnarray*}$$
So \(f_3\) is a homomorphism.

Answer 18

but \(f_1(g)= e_H=kernel~ f_1\)

Answer 19

Now we move to the next part.

Answer 20

Yes, I did the same for that part.

Answer 21

Well the second part is really easy to see. Where are you having trouble?

Answer 22

I want the smoothly logic from kernel f3 to intersection of kernel f1 and kernel f2.

Answer 23

by symbols, not verbal.

Answer 24

So, we start with the identity element in \(H \times K\) which is \((e_H, e_K)\). Consider the preimage of the identity element.

Answer 25

I mean I don't want to verbalize the logic.

Answer 26

\(f_3^{-1}(e_H,e_K) = kernel f_3\)

Answer 27

\(=\{g\in G~| f(g) =(e_H,e_K)\}\)

Answer 28

I mean f_3

Answer 29

I don't know how to avoid a bit of English, but it is clear that you end up with the set
\(\{g \in G : f_1(g) = e_H \wedge f_2(g) = e_K\}\)

Answer 30

Oh, so we just jump to this?? no need to put anything else between them?
Although I don't like to use verbal in Maths, but if it is needed, I can do it. :)

Answer 31

I think I got it. Thanks for your help.