Homomorphism and order of elements

• Nov 27th 2012, 06:14 PM
Piuzza12
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!
• Nov 27th 2012, 06:39 PM
jakncoke
Re: Homomorphism and order of elements
Yes.

Since |$\displaystyle \phi(G)$ | = n. That means, since $\displaystyle \phi(G)$ is a subgroup, then every element in that subgroup raised to the nth power is e.
so $\displaystyle (\phi(x))^n = e$ for $\displaystyle x \in G$. since $\displaystyle (\phi(x))^n = \phi(x^n) = e$ $\displaystyle x^n \in Ker(\phi)$
• Nov 28th 2012, 08:58 AM
Piuzza12
Re: Homomorphism and order of elements
Quote:

Originally Posted by jakncoke
$\displaystyle ... = \phi(x^n) = e$ $\displaystyle x^n \in Ker(\phi)$

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

Thank you!