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Thread: Proving ord(F(a)) divides order a

  1. #1
    Junior Member
    Oct 2009

    Post Proving ord(F(a)) divides order a

    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)
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  2. #2
    Senior Member
    Mar 2009
    Let $\displaystyle x \in G$ such that $\displaystyle x^n = e_{G}$ where $\displaystyle n \in \mathbb{N}$. Since $\displaystyle F$ is a group homomorphism we see that $\displaystyle F(x^n) = F(e_G) = e_{G'}$. On the other hand $\displaystyle F(x^n) = (F(x))^n$. Therefore, $\displaystyle (F(x))^n = e_{G'}$.
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