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