# Thread: Abstract Algebra Proof Homomorphism

1. ## Abstract Algebra Proof Homomorphism

Let G and H be two groups. If f: G $\rightarrow$ H is a homomorphism, a $\in$ G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let $e_{1}$ be the identity of G and $e_{2}$ be the identity of H)

I know $a^n$ = ?
and that $b^n$= ?

I have to prove from that given that n is a multiple of m.

Can I have some help with this proof? Thanks.

2. Originally Posted by Iceflash234
Let G and H be two groups. If f: G $\rightarrow$ H is a homomorphism, a $\in$ G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let $e_{1}$ be the identity of G and $e_{2}$ be the identity of H)

I know $a^n$ = ?
and that $b^n$= ?
Well, by definition of "order", you know that $a^n= e_1$ and $b^n= e_2$ (and there are no lower powers that give the identity).

I have to prove from that given that n is a multiple of m.

Can I have some help with this proof? Thanks.
Apply f to both sides of $a^n= e_1$. Then $f(a^n)= (f(a))^n= b^n= f(e_1)= e_2$. Now you know that $b^n= e_2$ and that m is the smallest integer such that $b^m= e_2$.

3. You lost me at the (f(a))^n = $b^n$. I know that f(a^n) = (f(a))^n but I don't see how that equals $b^n$ and then how that equals $f(e_1)$.

4. Originally Posted by Iceflash234
You lost me at the (f(a))^n = $b^n$. I know that f(a^n) = (f(a))^n but I don't see how that equals $b^n$ and then how that equals $f(e_1)$.
$f(a) = b$, so obviously, $(f(a))^n = b^n$. Now, $a^n = e_1 \Rightarrow f(a^n) = f(e_1) = e_2$

5. I understand that now. Thank you.