If $\varphi$ is a homomorphism, $\text{ord}(\varphi(a))$ divides $\text{ord}(a)$

Question

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 $$ \langle a\rangle=\{e,a,a^2, \dots, a^{m-1}\} $$ is a subgroup of $G$ of order $m$. Similarly: $$ \langle\varphi(a)\rangle=\{e,\varphi(a),\varphi^2(a), \dots, \varphi^{n-1}(a)\} \leq \varphi(G), \quad \text{ord}(\langle\varphi(a)\rangle)=n. $$

(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?

Answer 1

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

The homomorphism $\varphi$ induces a surjective homomorphism $\varphi'\colon\langle a\rangle\to\langle \varphi(a)\rangle$, because $\varphi(a^k)=\varphi(a)^k$. In particular, also $\varphi(a)$ has finite order. By the homomorphism theorem $$ \langle\varphi(a)\rangle\cong \langle a\rangle/\!\ker\varphi' $$ and Lagrange’s theorem applies to show that $|\langle\varphi(a)\rangle|$ divides $|\langle a\rangle|$.

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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 $f\colon\mathbb{Z}\to G$ defined by $f(k)=a^k$ has a kernel of the form $n\mathbb{Z}$ and the fact that $a^m=e$, but $a^k\ne e$ for $0<k<m$, implies $n=m$, so that $\langle a\rangle=\operatorname{im}f$ has exactly $m$ elements, being isomorphic to $\mathbb{Z}/m\mathbb{Z}$.

Answer 2

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 $\phi:G\rightarrow K$, then for any $H\leq G$, you can define the map $\phi|_H:H\rightarrow K$ by restriction and this is also a homomorphism and has kernel $\ker \phi \cap H$.

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

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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 $\langle a\rangle$ and $\langle \phi(a)\rangle$ are both cyclic groups. You can figure out that the only subgroups of a cyclic group $\langle a\rangle$ of order $n$ are those groups $\langle a^k\rangle$ where $k$ divides $n$. So the kernel of the map $\phi|_H:\langle a\rangle \rightarrow \langle \phi(a)\rangle$ is generated by some $a^k$ - which you can figure out is the minimal $k$ for which $\phi(a)^k=e$.

Answer 3

As I understand your question, you are trying to use the equation (1) to yield: $$\frac{|H|}{|\varphi(H)|} = s\cdot|ker(\varphi)|, \; s \in \mathbb{N}.$$ 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 $s \in \mathbb{N}$.