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 *x*^{n} 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 x^{n} must be mapped to the neutral element of H. Is that correct?

Any help is greatly appreciated!

Thanks!

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) $

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!