The Order Of A Is A Multiple Of The Order Of FA Or AKFA Shaughn

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Problem 89 (The order of $a$ is a multiple of the order of $f(a)$, or $|a|=k|f(a)|$)

Suppose that $f:G\to H$ is a homomorphism. Prove that if $a\in G$ has order $n$, show that $|f(a)|$ divides $n$, i.e. show that $|a|$ is a multiple of $|f(a)|$.

Solution

Let $a\in G$ with $|a|=n$. We compute $(f(a))^n=f(a^n)=f(e_G)=e_H$. It was shown in a previous exercise that if $a^m=e$, then the order of $a$ divides $m$. Since $(f(a))^n=e_H$, we conclude that $|f(a)|$ divides $n$.

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