# Thread: Homomorphism and order of elements

1. ## Homomorphism and order of elements

Hi,

I was wondering if someone could help me with this question I have.

"If f: G--> H is a homomorphism, prove the following. "If the range of f has n elements, then xn is in the kern f for every x in G. ""

I know that the kern f is a set K that has all elements of G that are carried by f onto the neutral element in H.

I also know that the range of f is a subgroup of H.

So what the question is asking to prove is that if we take every element of G and raise it to the power n where n is the number of elements in the range, then all of those xn must be mapped to the neutral element of H. Is that correct?

Any help is greatly appreciated!

Thanks!

2. ## Re: Homomorphism and order of elements

Yes.

Since | $\phi(G)$ | = n. That means, since $\phi(G)$ is a subgroup, then every element in that subgroup raised to the nth power is e.
so $(\phi(x))^n = e$ for $x \in G$. since $(\phi(x))^n = \phi(x^n) = e$ $x^n \in Ker(\phi)$

3. ## Re: Homomorphism and order of elements

Originally Posted by jakncoke
$... = \phi(x^n) = e$ $x^n \in Ker(\phi)$
I have another question. How you you get from $\phi(x^n) = e$ $x^n \in Ker(\phi)$?

Thank you!