# Prove isomorphism

• Apr 12th 2010, 07:41 AM
rainyice
Prove isomorphism
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.
• Apr 12th 2010, 03:29 PM
Drexel28
Quote:

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 $|\theta(g)|=n$
• Apr 12th 2010, 05:04 PM
rainyice
Quote:

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 ..... ><"
• Apr 12th 2010, 08:51 PM
Drexel28
Quote:

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

If $g^n=e$ then $|g|\mid n$
• Apr 13th 2010, 07:37 PM
MissMousey
Quote:

Originally Posted by Drexel28
If $g^n=e$ then $|g|\mid n$

That's if we assume that group G is finite. However, can we assume that these two groups are finite?
• Apr 13th 2010, 07:46 PM
Drexel28
Quote:

Originally Posted by MissMousey
That's if we assume that group G is finite. However, can we assume that these two groups are finite?

Assume that $|\theta(g)|=\infty$ but $|g|=n<\infty$ then $\theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so $|\theta(g)|\leqslant n<\infty$, contradiction. You can do the reverse just as easily.