Jtdylktuy

Assume $varphi: , G longrightarrow G'$ is a homomorphism and $a in G$ an element of finite order.
If $m=text{ord}(a)$ and $n=text{ord}(varphi(a))$, show that $n big| m$.

To begin with, it's known that
$$
langle arangle={e,a,a^2, dots, a^{m-1}}
$$
is a subgroup of $G$ of order $m$. Similarly:
$$
langlevarphi(a)rangle={e,varphi(a),varphi^2(a), dots, varphi^{n-1}(a)} leq varphi(G), quad text{ord}(langlevarphi(a)rangle)=n.
$$

(Given that, it is sufficient to show that for every $H leq G$, $|varphi(H)|$ divides $|H|$.)

The Fundamental Theorem of Homomorphisms gives:
$$
G/text{ker} varphi longrightarrow varphi (G) iff |G|stackrel{(1)}{=}|text{ker} varphi| |varphi(G)|
$$

Having all these in mind, I'm trying to use equation $(1)$ to yield:
$$
frac{|H|}{|varphi(H)|}=s cdot |text{ker} varphi|, quad s in mathbb{N}
$$
using Langange's Theorem. Any hints?

An element $ain G$ has finite order $n$ if and only if $langle arangle={a^k:kinmathbb{Z}}$ (which is a subgroup of $G$) has $n$ elements. (*)

The homomorphism $varphi$ induces a surjective homomorphism $varphi'colonlangle arangletolangle varphi(a)rangle$, because $varphi(a^k)=varphi(a)^k$. In particular, also $varphi(a)$ has finite order. By the homomorphism theorem
$$
langlevarphi(a)ranglecong langle arangle/!kervarphi'
$$
and Lagrange’s theorem applies to show that $|langlevarphi(a)rangle|$ divides $|langle arangle|$.

The statement (*) is true either by definition or, in case the order is defined as the minimal positive integer $m$ such that $a^m=e$, by observing that the homomorphism $fcolonmathbb{Z}to G$ defined by $f(k)=a^k$ has a kernel of the form $nmathbb{Z}$ and the fact that $a^m=e$, but $a^kne e$ for $0<k<m$, implies $n=m$, so that $langle arangle=operatorname{im}f$ has exactly $m$ elements, being isomorphic to $mathbb{Z}/mmathbb{Z}$.

As I understand your question, you are trying to use the equation (1) to yield:
$$frac{|H|}{|varphi(H)|} = scdot|ker(varphi)|, ; s in mathbb{N}.$$ As in the question, $varphi$ is a group homomorphism which is not restricted to $H$, hence the equation you aim to prove doesn't necessarily hold. In fact if $G=S_3$, $G'={e'}$ and $varphi$ is the trivial homomorphism. For $H={{e}}$, $varphi{H}={{e'}}$ and thus $|H|= 1$, $|varphi(H)|=1$, $|ker(varphi)|= 6 $ which is not satisfied for any $s in mathbb{N}$.

Equation (1) basically just gives you the answer. Let $H$ be any subgroup of $G$. Then, if you apply that equation to the restriction of $phi$ to $H$, you get
$$H/(ker phi cap H)cong phi(H)$$
Thus, you must have
$$|H|=|phi(H)|cdot |ker phi cap H|$$
So $|phi(H)|$ divides $|H|$. They key insight is that if you have a homomorphism $phi:Grightarrow K$, then for any $Hleq G$, you can define the map $phi|_H:Hrightarrow K$ by restriction and this is also a homomorphism and has kernel $ker phi cap H$.

You would still need to show that $langle phi(a)rangle = phi(langle arangle)$ to finish the proof, though this isn't too hard, depending on how you defined these terms.

This is a nice proof of the fact, but it's worth noting that there is a simpler more common proof based on the following statement:

$a^k=e$ if and only if $operatorname{ord}(a)|k$.

This is sometimes used as the definition of order, although it would also be a theorem under other definitions.

Then, you note that $a^{operatorname{ord}(a)}=e$, so $phi(a)^{operatorname{ord}(a)}=e$ so $operatorname{ord}(phi(a))|operatorname{ord}(a)$.

Better yet is to note that this proof is really what's going on behind the scenes of your proof: you can find that $langle arangle$ and $langle phi(a)rangle$ are both cyclic groups. You can figure out that the only subgroups of a cyclic group $langle arangle$ of order $n$ are those groups $langle a^krangle$ where $k$ divides $n$. So the kernel of the map $phi|_H:langle arangle rightarrow langle phi(a)rangle$ is generated by some $a^k$ - which you can figure out is the minimal $k$ for which $phi(a)^k=e$.