Homomorphisms, finite groups, and primes

kathrynag
Messages
595
Reaction score
0

Homework Statement


1. Let G and H be finite groups and let a: G → H be a group homomorphism. Show
that if |G| is a prime, then a is either one-to-one or the trivial homomorphism.
2. Let G and H be finite groups and let a : G → H be a group homomorphism. Show
that if |H| is a prime, then a is either onto or the trivial homomorphism.


Homework Equations





The Attempt at a Solution


1. We know a(b)a(c)=a(bc) since it is a homomorphism
order is prime.
need to show a(x1)=a(x2) implies x1=x2. I'm confused on how the oder being prime plays into this.
 
Physics news on Phys.org
if |G| is prime then it is a cyclic group generated by one single element. Hope this helps.

hint: write f for your homomorphism and not a.
 
You don't need to mess around with elements. Hint:

The kernel of a homomorphism from G to H is a _______ of G
The image of a homomorphism from G to H is a _______ of H
 
hmm, may i ask what trivial homomorphism means?
 
The homomorphism which maps everything to 0.
 
Since it's a cyclic group generated by one element it must be one to one since there is only element.

|H| is prime. H is a cyclic group generated by one element. Must have x such that f(x)=y
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
13
Views
562
Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
429
Replies
8
Views
2K
Back
Top