Question about homomorphism of cyclic group if $\varphi: G\to H$ is homomorphism.
How do I prove that if $a\in G$ have finite order so $\varphi(a)$ had finite order to, and that:$$ord(\varphi(a))\mid ord(a)$$
Thank you!
 A: Since $$φ(a)^n=φ(a^n)=φ(e)=e$$ we can use this to show that the order of $φ(a)$ must divide $n$. Let $m$ denote the order of $φ(a)$. Then $n=qm+r,\ q\in\Bbb N,\ r<m$. Now $$e=φ(a)^n=\left(φ(a)^m\right)^q*φ(a)^r=e*φ(a)^r=φ(a)^r$$
Since $m$ is the smallest integer such that $φ(a)^m=e$ and $r<m$ we conclude that $r=0$, hence $n=qm$.
A: Note that a homomorphism $\varphi$ must satisfy $\varphi(g*h)=\varphi(g)\cdot\varphi(h)$ for all $g,h\in G$.  Let $e\in G$ be the identity.  Then $\varphi(g)\cdot \varphi(e)=\varphi(g*e)=\varphi(g)=\varphi(e*g)=\varphi(e)\cdot\varphi(g)$, so $\varphi(e)$ is the identity in $H$.  This also implies $\varphi(a^n)=\varphi(a)^n$, so $$\varphi(e)=\varphi(a^{\text{ord}(a)})=\varphi(a)^{\text{ord(a)}}$$ Therefore $\text{ord}_H(\varphi(a))|\text{ord}_G(a)$
A: If $\def\ord{\operatorname{ord}}n=\ord(a)$ then $e_H=\phi(e_G)=\phi(a^n)=\phi(a)^n$. Therefore $\ord(\phi(a))$ divides $n$ (the only powers of an element that give the identity are the multiples of its order).
A: Saying that $a$ has finite order means that 
$$
\langle a\rangle=\{a^n:n\in\mathbb{Z}\}
$$
is a finite subgroup of $G$. Moreover the order of $a$ is exactly $|\langle a\rangle|$. The homomorphism $\varphi\colon G\to H$ induces a surjective homomorphism
$$
\phi'\colon \langle a\rangle\to\langle \varphi(a)\rangle
$$
because $\phi(a^k)=(\phi(a))^k$ for each integer $k$.
Thus the order of $\varphi(a)$ is finite and from the homomorphism theorem you get that
$$
\langle \varphi(a)\rangle\cong\langle a\rangle/\ker\varphi'
$$
so the order of $\langle \varphi(a)\rangle$ divides the order of $\langle a\rangle$, which is what you wanted to prove.
