Prove isomorphism

**rainyice** · April 12th 2010, 07:41 AM

If G and H are groups and f:G --> H is an isomorphism, prove that a and f(a) have the same order, for any a E G

I understand the concept. Isomorphism has one to one property, hence a have the same order of f(a) ... however, I don't not know how to prove it mathematically.

**Drexel28** · April 12th 2010, 03:29 PM

Originally Posted by rainyice

If G and H are groups and f:G --> H is an isomorphism, prove that a and f(a) have the same order, for any a E G

I understand the concept. Isomorphism has one to one property, hence a have the same order of f(a) ... however, I don't not know how to prove it mathematically.

Notice that if $|g|=n$ then $\theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so from basic group theory $|\theta(g)|\mid n$ . But! Using the exact same idea and noting that $\theta^{-1}:H\to G$ is an isomorphism we note that $n\mid|\theta(g)|$ and so $|theta(g)|=\pm n$ but since $|\theta(g)|\geqslant 0$ it easily follows that $|\theta(g)|=n$

**rainyice** · April 12th 2010, 05:04 PM

Originally Posted by Drexel28

Notice that if $|g|=n$ then $\theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so from basic group theory $|\theta(g)|\mid n$ . But! Using the exact same idea and noting that $\theta^{-1}:H\to G$ is an isomorphism we note that $n\mid|\theta(g)|$ and so $|theta(g)|=\pm n$ but since $|\theta(g)|\geqslant 0$ it easily follows that $|\theta(g)|=n$

I don't get where the |theta(g)| \ n comes from ..... ><"