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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.

MJD
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Cornman
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  • This somewhat related question seems to imply that we could say instead that $\operatorname{ord}(g) = 0$ in which $a\mid 0$ is always true, or keep the convention that $\operatorname{ord}(g) = \infty$ and then $a\mid \infty$ by convention as well. Either one would be fine if $g$ has “infinite” order, but the second option is more common and has the advantage that $\operatorname{ord}(g) = \operatorname{card}{\langle g\rangle}$. – shoteyes Jun 13 '21 at 05:38

1 Answers1

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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.

Cooler Paradox
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