# Thread: Abstract Algebra Proof Homomorphism

1. ## Abstract Algebra Proof Homomorphism

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

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

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

I know $\displaystyle a^n$ = ?
and that $\displaystyle b^n$= ?
Well, by definition of "order", you know that $\displaystyle a^n= e_1$ and $\displaystyle b^n= e_2$ (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 $\displaystyle a^n= e_1$. Then $\displaystyle f(a^n)= (f(a))^n= b^n= f(e_1)= e_2$. Now you know that $\displaystyle b^n= e_2$ and that m is the smallest integer such that $\displaystyle b^m= e_2$.

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

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

5. I understand that now. Thank you.