
Homomorphism and periods
Let $\displaystyle f : G > G'$ be a homomorphism. Let $\displaystyle a$ be an element of $\displaystyle G$ of period 10. Prove that $\displaystyle f(a)$ must have period 1 or 2 or 5 or 10.
I can prove that $\displaystyle f(a)$ may have period 10.
$\displaystyle a^{10} = e$ because $\displaystyle a$ have period 10
$\displaystyle f(a^{10}) = [f(a)]^{10} = e'$ by property of homomorphism
Therefore $\displaystyle f(a)$ may have period 10.

Suppose that $\displaystyle f(a)$ has period $\displaystyle k$. Then by the division algorithm, we can write $\displaystyle 10=qk+r$ where $\displaystyle q,r$ are integers and $\displaystyle 0\leq r<k1$. Then, following your original calculation,
$\displaystyle e'=f(a^{10})=f(a^{qk+r})=f(a^{qk})f(a^r)=[f(a)]^{qk}[f(a)]^r$
First note that $\displaystyle [f(a)]^{qk}=e'$ as $\displaystyle k$ was the period. This means that $\displaystyle [f(a)]^r=e'$. But since $\displaystyle r<k$, we must take $\displaystyle r=0$ (i.e. $\displaystyle k$ was the smallest nonzero number such that $\displaystyle [f(a)]^k=e'$). This means that $\displaystyle 10=qk$, so the period $\displaystyle k$ must divide 10. So we can conclude that $\displaystyle k=1, 2, 5,\text{ or } 10$.