Proving ord(F(a)) divides order a

**ChrisBickle** · December 6th 2009, 05:12 PM

Allright here is a problem from one of my hw assignments and i have no idea how to solve it. Im sure there is probably a theorem or something that can be used to make it quite easy but i cant think of anything.

Let F: G->G' be a group homomorphism and suppose a is in G and a has finite order. Prove that the ord(F(a)) divides ord(a)

**lvleph** · December 6th 2009, 05:30 PM

Let $x \in G$ such that $x^n = e_{G}$ where $n \in \mathbb{N}$ . Since $F$ is a group homomorphism we see that $F(x^n) = F(e_G) = e_{G'}$ . On the other hand $F(x^n) = (F(x))^n$ . Therefore, $(F(x))^n = e_{G'}$ .

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