Abstract Algebra Proof Homomorphism

**Iceflash234** · May 2nd 2010, 07:46 AM

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.

**HallsofIvy** · May 2nd 2010, 07:55 AM

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 $b^m= e_2$ .