Is $\{k \in \mathbb{Z}_{40} \,|\, k \text{ is a multiple of 3}\}$ a subgroup of $\mathbb{Z}_{40}$? Is $S = \{k \in \mathbb{Z}_{40} \,|\, k \text{ is a multiple of 3}\}$ a subgroup of $\mathbb{Z}_{40}$?
No, right? I have a student who doesn't believe my argument of non-closure because $3 \in S$ and $39 \in S$ but $3 + 39 \mod 40 \equiv 2 \not\in S$ since $2$ isn't a multiple of $3$.
The student claims that this is the group $3\mathbb{Z}_{40}$ and that $3\mathbb{Z}_{40} = \mathbb{Z}_{40}$ because every multiple of 3 loops around and that an algebra professor told her this. But it's not making any sense. I'm thinking either she or the professor are thinking of the cyclic subgroup generated by $3$, $\langle 3 \rangle = \mathbb{Z}_{40}$.
EDIT: Earlier in the student's problem set, we have a similar problem that asks you if $T = \{k \in \mathbb{Z}_{40} \,|\,k \text{ is odd}\}$ is a subgroup of $\mathbb{Z}_{40}$. The counter example I used was that $1 + 3 = 4 \not\in T$. Would this likely imply that my assumption of the definition was what the problem set had in mind? Since technically you'd have a loop-around of odd numbers becoming even.
 A: The question is actually ambiguous, and I would be rather disappointed that the professor did not realise this.
The point is that the statement

$a$ is a multiple of $b$

by itself, is ambiguous.  Unless otherwise specified, it means that

there is an integer $c$ such that $a=bc$.

For example, $24$ is a multiple of $8$ because $24=8\times3$ and $3$ is an integer.
However, there is no need to be talking about integers!  Suppose that we are talking about even numbers only.  Then

$a$ is a multiple of $b$

means

there is an even integer $c$ such that $a=bc$,

and in this case $24$ is not a multiple of $8$.
Usually this ambiguity is unimportant, because the only things we could be talking about are the integers.  However, in your question, there is another alternative: we could be talking about ${\Bbb Z}_{40}$.


*

*If we are talking about divisibility in $\Bbb Z$: then
$$\{k\in{\Bbb Z}_{40}\mid k\ \hbox{is a multiple of $3$}\}
  =\{0,3,6,\ldots,39\}\ ;$$
this is not a subgroup and you are right.

*If we are talking about divisibility in ${\Bbb Z}_{40}$, then
$$\eqalign{
  \{k\in{\Bbb Z}_{40}\mid k\ \hbox{is a multiple of $3$}\}
  &=\{k\in{\Bbb Z}_{40}\mid 3x\equiv k\pmod{40}\ \hbox{has a solution}\}\cr
  &={\Bbb Z}_{40}\ ;\cr}$$
this is (obviously!) a subgroup, and the others are right.

A: I think the problem might be that your notion of divisibility is not well-defined.
Recall that $\mathbb Z _{40} =\mathbb Z/40\mathbb Z$. The elements are not actually integers, but equivalence classes of integers. You argue that $3 \in S$ because $3$ is divisible by $3$, but we are really talking about $\bar 3$. However, we know that $\overline {43} = \bar 3$, but $43$ is not divisible by $3$.
Since your divisibility condition depends on the representative of the equivalence class, it isn't well-defined.
Instead, you might say that $m \in \mathbb Z$ is a multiple of $3$ if there exists some $z \in \mathbb Z$ such that $z\cdot 3 = m$. 
A: Your student is correct.
A multiple of a quantity is a product of the quantity with any integer.
So,
$S = \{ k\in \mathbb{Z}_{40}|k$ $is$ $a$ $multiple$ $of$ $3 \} = \{ k\in \mathbb{Z}_{40}| k=3a, a\in \mathbb{Z}\} = \mathbb{Z}_{40}$.
Hence, $S$ is a subgroup of $\mathbb{Z}_{40}$.
