Is there a convention for divisibilty $a\mid \infty$? The following problem:
Let $G,H$ be groups and $\varphi: G\to H$ be a group homomorphism. Prove that $\mathrm{ord}(\varphi(g))\mid\operatorname{ord}(g)$.
This is easy when $\operatorname{ord}(g)=n<\infty$.
But what if $\operatorname{ord}(g)=\infty$. Is there a convention that $a\mid \infty$ always holds, or is this task just incomplete?
Thanks in advance.
 A: The standard definition of $a |  b$ is that $a$ and $b$ are integers, and there is an integer $c$ for which $b = ac$. When we are dealing with sets of infinite cardinality, this definition has no meaning. However, there is a natural way to extend this notion.
Namely, if $a | b$, this means that $b$, thought of as a set of $b$ elements, can be partitioned in such a way that each set in the partition has the same size $a$. If $b = ac$ as above, then the number of sets in the partition is $c$. We can write this as
$\{1, 2, \ldots, b\} = \bigsqcup_{i = 1}^c A_i = \bigsqcup_{i = 1}^c \{x_{i1}, x_{i2}, \ldots, x_{ia}\}$.
In particular, $|A_i| = a$ for all $i = 1, 2, \ldots, c$. This idea then extends immediately to infinite sets, since infinite sets can also be partitioned.
For an example of how this would apply to a group, consider the group $G = (\mathbb{Q} \setminus \{0\}, \cdot)$ of non-zero rational numbers under multiplication. Then $H = \{-1, 1\}$ is a subgroup of $G$. Clearly, $|H| = 2$, but $G$ is infinite. However, it still makes sense to think of the order of $H$ as "dividing" the order of $G$ in this more general sense of partitioning into sets of equal size. That is, the set of cosets $P = \{gH \, | \, g \in G\}$ is a partition of $G$, and each set in $P$ has the same size, namely $|H| = 2$. There are infinitely many sets in the partition $P$, which are in natural correspondence with the positive rationals. That is,
$P = \{\{-q, q\} \, | \, q \in \mathbb{Q}, q > 0\}$.
Long story short, dividing means partitioning into sets of equal cardinality.
