Question

a)Prove that a nonzero homomorphic image of a divisible group must also be divisible.
b) Prove that a nonzero finite abelian group cannot be divisible.
Please help

Answer 1

@Alchemista

Answer 2

fiber of h is just g? f is homomorphism, not one to one. Can we conclude that only 1 g in G is preimage of h?

Answer 3

a) Let \(G\) be a divisible group and take an arbitrary homomorphism \(f\) such that \(H=f(G)\) is not trivial.

We want to show that \(H\) is divisible. Consider any \(n \in \mathbb{N}\) and any \(h \in H\) we want to show that there is some \(x \in H\) such that \(nx = h\).

Let \(g \in f^{-1}(h)\) and by the divisibility of G we can find some \(y \in G\) such that \(ny = g\). Then we have that \(f(ny) = f(g) = h = n \cdot f(y)\) since \(f\) is a homomorphism.

Therefore let \(x = f(y)\) and we have an \(x\) such that \(nx = h\).

Note: \(f(ny) = n \cdot f(y)\) since we have \(f(y+y+...+y)=f(y)+f(y)+...+f(y)\) since \(f\) is a homomorphism.

Answer 4

There, any more objections?

Answer 5

Yes, It 's good.
Can we do directly from G?
Let \(g\in G\), there exist \(g'\in G\) such that \(ng'=g\)
\(f:G\rightarrow H\\~~~~~~ g\rightarrow f(g)\\~~~~~~ng'\rightarrow f(ng')= nf(g')\)

Answer 6

But you are trying to show that \(H\) is divisible.

Answer 7

That shows H is divisible, right?

Answer 8

Let me check

Answer 9

I'm not sure that follows. I think it is better to prove that \(H\) is divisible based on the definition of divisibility for \(H\).

Answer 10

Oh, I got you :) since we need prove that every element in H is the form of nh' to get H is divisible. So we must start from H. right?

Answer 11

You need to show that for every \(n \in \mathbb{N}\) and every \(h\in H\) that there is some \(x \in H\) such that \(nx = h\).

Answer 12

Yes, :)
How about b)

Answer 13

Have you thought about it yet?

Answer 14

I confused about it. Since we have Q (rational group) is divisible, but it is infinite. How finite one can limit the divisibility?

Answer 15

One more think, abelian group indicate to +operator in group in this case and we know that all modulo Z is abelian and they are finite. I mean \(\mathbb Z_n\) is divisible and \(\mathbb Z_n/\mathbb Z_2\) is divisible, right?

Answer 16

Did you learn about the fundamental theorem of finite abelian groups?

Answer 17

Yes, I did

Answer 18

Then you should know that all such groups are isomorphic to a direct product of cyclic groups of prime power.

Answer 19

So take an arbitrary direct product of cyclic groups (of prime power) and show that there is an \(n \in \mathbb{N}\) and a \(g\) in that arbitrary group you picked such that there is no \(y\) in that group such that \(ny = g\)

Answer 20

I think .... I get the idea. Let me try. :)

Answer 21

Consider \(\mathbb Z_6\). Let \(A=\{[0],[3]\}\) and \(B=\{[0],[2],[4]\}\)
then \(\mathbb Z_6= A+B\)

Answer 22

order of A is 2 , order of B is 3, then they are relative prime also. then?

Answer 23

4 = 2*2 , how to argue more?

Answer 24

but there is no element \(x\) in A+B, such that n x=3

Answer 25

What did you choose as \(n\)?

Answer 26

by definition of divisibility, n is "for all"

Answer 27

That is if n =1, then [3] =[3] in A+B
n =2 , then [3]+[3]=[0] in A+B
and so on

Answer 28

To show a group is NOT divisible, you need to provide an \(n \in \mathbb{N}\) and a \(g\) in the group such that the following equation has no solution:
\(x^n = g\)

Answer 29

I do confuse. why not \(xn \) but \(x^n\)

Answer 30

It is just different notation for the same thing.

Answer 31

In abelian groups you often use additive notation.

Answer 32

In other words \(g^n = g\cdot g ... \cdot g \) is written as \(g + g + ... + g\) or \(ng\)

Answer 33

If I pick g = 3, then I need some g' in G that is either 0,2,4 (let say 2) and particular n such that 2^n = 3, right?
but there is no such n in N,
hence the group is not divisible, right?

Answer 34

No, you pick \(n\) and \(g\) and show there is no \(y\)

Answer 35

:(

Answer 36

nope, any element in \(\mathbb Z_6\) is cyclic, so that \(g^n \in \mathbb Z_6\)

Answer 37

for example, if I pick 3 in the group, then for all n, I have 3^n in the group.

Answer 38

3^2 =6 =0
3^3 =9=3
3^4= 81=3 so on

Answer 39

I told you, \(n\) and \(g\) are constants. You need to pick \(n\) and \(g\) and show there is no \(x\) such that \(nx = g\). You are not showing that there is no \(n\).......

Answer 40

What is g? is it an element in G?

Answer 41

Yes

Answer 42

so, now I pick 2

Answer 43

You need to pick a natural number and an element in the group.

Answer 44

if element in group is 0, then it is so trivial. I don't want the special case. It confuses me.
Now I pick element 2 in the group, and pick constant n = 3, then there is no other element among the rest (0,3,4) such that n * that element =2, right?

Answer 45

Think about this: take \(\mathbb{Z}_p\) and try \(n = p\) and \(g = 1\)

Can you solve \(px = 1\)?

Answer 46

No

Answer 47

So for \(\mathbb{Z}_{p_1} \times \mathbb{Z}_{p_2} \times ... \times \mathbb{Z}_{p_m}\):

Can you solve \((p_1p_2...p_m)x = 1\)

Answer 48

No

Answer 49

And all finite abelian groups are of the above form.

Answer 50

:)
Thanks a ton. I got it.
Thanks for being patient to me.

Answer 51

That is what I want :). It is in general , not just particular case. Since the problem is Prove that .......
I think it must be true for any case and we need start from the general one.

I am so sorry for my slowness and stubbornness. ;)

Answer 52

The idea is to start with a simple case and then try to generalize it.

Answer 53

This is almost always how you try to solve problems. Start simple and then try to make it work for all cases.

Answer 54

Group theory is sooooooooo hard to me. I learn it hardly. Just one line of theorem, I spent all day to figure out what it is, how to apply to the problem. ha!!

Answer 55

Do you like topology/analysis more?

Answer 56

Actually, I don't!! but whenever I get it, I am sooooooooo happy.

Answer 57

I like it more than algebra.

Answer 58

The feeling of winning the problem is awesome. But sometimes, I just want to give up.

Answer 59

Where is your school?

Answer 60

Canada

Answer 61

Did you take community college before University?

Answer 62

no

Answer 63

That is why you are that good.

Answer 64

I took community college in 2 years. I was one of the best students there. When I move to University, my classmates who didn't take Community College know the things I NEVER know before.

Answer 65

I works 3 times harder than them to get back what I don't know but I am not good as them still, :(

Answer 66

One way to get ahead is to solve problems from the text even if they are not in the suggested problems list. The only way to get better is to solve lots of problems.

Answer 67

Yes, I know but I don't have enough time to do that. Go over the lecture and assignment and review and fix the missing things take all my time. :)