# Prove only group homomorphism between Z5 and Z7 is the trivial one.

1. Aug 22, 2013

### mahler1

1. The problem statement, all variables and given/known data.

Prove that the only homomorphism between Z5 and Z7 is ψ(x)=0 (the trivial homomorphism).

3. The attempt at a solution.

I wanted to check if my solution is correct, so here it goes:

Any element x in Z5 belongs to the set {0,1,2,3,4}

So, I trivially start by saying that ψ(1)=ψ(1) and then, using the fact that ψ(x+y)=ψ(x)+ψ(y), I get:
ψ(2)=ψ(1+1)=ψ(1)+ψ(1)=2ψ(1)
.
.
.
ψ(5)=ψ(1+1+1+1+1)=ψ(1)+ψ(1)+ψ(1)+ψ(1)+ψ(1)=5ψ(1)

But 5=0 (mod 5) and, as ψ is a homomorphism, it must be 0= ψ(0)=ψ(5).
By transitivity, 0=5ψ(1) (mod 7) but 5≠0 (mod 7), then ψ(1)=0. But ψ(x)=xψ(1)=0 for all x in Z5, hence Hom(Z5,Z7)=0.

2. Aug 22, 2013

### Office_Shredder

Staff Emeritus
I think this last line needs to be spelled out a little more, essentially you have claimed that 5x = 0, and since 5 is not equal to 0 that x is equal to 0. There exist rings where you can multiply two nonzero numbers and get a zero, so you should probably explicitly prove that you're OK in Z7 unless this is something you've already done in your class

3. Aug 22, 2013

### Zondrina

Hint : You need to use the fact that :

$gcd(|\mathbb{Z}_5|, |\mathbb{Z}_7|) = 1$

as well as Lagrange's theorem.

Last edited: Aug 23, 2013
4. Aug 24, 2013

### mahler1

Yes, I see what you mean. If I am in Z6, 3≠0 (mod 6) but from 3x=0 (mod 6) I can't conclude that x=0 (mod 6) as any x even would satisfy 3x=0 (mod 6). In the original problem I could say x=0 (mod 7) just because gcm(5,7)=1. I am not so sure how to prove what I've claimed, I'll check it. Thanks.

5. Aug 25, 2013

### mathwonk

prove: the only factorization of 5 as a product of a factor of 5 and a factor of 7 is 5.1.