${\rm order}(x)$ is infinite or ${\rm order}(h(x))$ divides ${\rm order}(x)$ even if we don't assume ${\rm order}(h(x))$ is finite right?

Question

Let $A,B$ be groups with a homomorphism $h: A \to B$. Let $x,y \in A$.

I'm supposed to show that if $h(x)$ has finite order in $B$, then ${\rm order}(x)$ is infinite or ${\rm order}(h(x))$ divides ${\rm order}(x)=n < \infty$.

And of course the answer is

• https://proofwiki.org/wiki/Element_to_Power_of_Multiple_of_Order_is_Identity

• https://proofwiki.org/wiki/Order_of_Homomorphic_Image_of_Group_Element

What I'm wondering is the relevance of the assumption '$h(x)$ has finite order in $B$'. I think the only role it serves to make '${\rm order}(h(x))$' well-defined because actually I think the statement

${\rm order}(x)$ is infinite or ${\rm order}(h(x))$ divides ${\rm order}(x)$

is ('sensible' and) true because ${\rm order}(x) < \infty \implies{\rm order}(h(x)) < \infty$.

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Questions:

1. Is the statement "${\rm order}(x)$ is infinite or ${\rm order}(h(x))$ divides ${\rm order}(x)$" true?

2. If not, then why not? If so, then am I right about the role of the assumption that ${\rm order}(h(x)) < \infty$ ?

3. Can we just say instead like

Show that if ${\rm order}(x) < \infty$, then

(A) ${\rm order}(h(x)) < \infty$ because ​${\rm order}(h(x)) \le{\rm order}(x)$

• (the $(h(x))^n = h(x^n=1_A)=1_B$ part)

(B) ${\rm order}(h(x))$ divides ${\rm order}(x)$

• (the division algorithm part)

?