# Trivial group homomorphism from G to Q

1. Jan 21, 2016

### tomkoolen

• Member warned about posting without the template
Hello, I have to solve the following problem:

Show that a homomorphism from a finite group G to Q, the additive group of rational numbers is trivial, so for every g of G, f(g) = 0.

My work so far:

f(x+y) = f(x)+f(y)
I know that |G| = |ker(f)||Im(f)|

I think that somehow I have to find that Im(f) = 1 but I don't know how. Can anybody help me please?

2. Jan 21, 2016

### Samy_A

You could use the fact that every element of G has a finite period.

3. Jan 21, 2016

### HallsofIvy

Staff Emeritus
The only finite subgroup of the additive group of rational numbers is {0}. (NOT {1}!)

4. Jan 21, 2016

### tomkoolen

Okay I understand that, but now I only know that there is an isomorphism to Z/nZ with n the number of elements. Then I know that f is one-to-one, so |ker(f)| = 1 and Im(f) would equal n, but that's not what I need. Can you give me one more hint?

5. Jan 21, 2016

### Samy_A

Aren't you making this too complicated?
Take an element $g \in G$. Then $g^p=e$ for some positive integer $p$, ( $e$ is the identity of G) . What information can you glean from $f(g^p)=f(e)$?

6. Jan 21, 2016

### tomkoolen

f(g^p) = f(e) so that is e of Q (= 0). And you can conclude that order(f(x)) is finite as well and then you can conclude that because {0} is the only finite subgroup of Q all elements of G will be mapped to 0?

Thanks for the help!

7. Jan 21, 2016

### HallsofIvy

Staff Emeritus
I wasn't thinking that complicated! I would just use the fact that a homomorphism from group G to group H maps G to a subgroup of H.

8. Jan 21, 2016

### Staff: Mentor

Isn't this the same amount of work, sorry: line of proof? Because with that you have to prove your statement in #3.

9. Jan 21, 2016

### nuuskur

What are the requirements for f to be a group homomorphism?